Physics 21 Solutions

March 21, 2018 | Author: Oğuzhan Odbay | Category: Capacitor, Flux, Electrical Resistivity And Conductivity, Force, Electric Field

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Physics 21 Fall, 2011

Equation Sheet

speed of light in vacuo Gravitational constant Avogadro’s Number Boltzmann’s constant charge on electron free space permittivity free space permeability gravitational acceleration

F2 on 1 =

3.00 × 108 m/s 6.67 × 10−11 N m2 /kg2 6.02 × 1023 mol−1 1.38 × 10−23 J/K 1.60 × 10−19 C 8.85 × 10−12 C2 /(N m2 ) 4π × 10−7 T m/A 9.807 m/s2

c G NA kB e 0 µ0 g

1 q1 q2 (r1 − r2 ) 4π0 |r1 − r2 |3

λ σ ; Eplane = r 20 σ for plate 0 capacitor A A Q = CV ; C = 0 K = d d 2 Q 2 Ucap = 12 CV = 12 C Uind = 12 LI 2 ρL V = IR R= A P = IV P = I 2R 1 2π0 Q E= = 0 A

1 dQ (r − r ) 4π0 |r − r |3

E = −∇V ∂V ∂V ˆ ∂V = − î + ˆj +k ∂x ∂y ∂z V f − Vi = −

f

E · dl

i

1 Q dQ 1 ; dV = 4π0 r 4π0 |r − r | 1 uelec = 12 0 E 2 , umag = B2 2µ0 V=

Work =

R = mv⊥ /(qB)

A × B =

F · dl

E · dA =

Q 0

d dt

E · dl = −

u du = a2 + u 2

a2 + u 2

du 1 = tan−1 a2 + u 2 a

v=

T /ρ

u a

ln a2 + u

1 2

(T =tension)

v = (347.4 m/s)

T /300

v = λf = ω/k ω = 2πf

k = 2π/λ

T = 1/f

(T =period)

P = 12 ρA2 ω 2 v

2

1 ωC

RC time constant = RC LR time constant = L/R Q(t) for RLC circuit Q0 exp(−Rt/2L) cos ωt ω2 =

= sin θ cos

π 2

1 R2 − LC 4L2

± cos θ sin

E · dA

u du = a2 + u 2

XR = R, XL = ωL, XC =

solenoid B = µ0 nI solenoid L = µ0 N 2 A/l

C = 2πr C = πd A = πr2 A = 4πr2 V = 43 πr3

π 2

circumference of circle circumference of circle area of circle surface area of sphere volume of sphere

cos(a ± b) = cos a cos b ∓ sin a sin b

du √ = ln u + a2 + u2 2 2 a +u

π ) 2

long wire: B =

ξeffective = ξ1 + ξ2

= ± cos θ

B · dA

d B · dl = µ0 I + µ0 0 dt

√

sin(θ ±

µ0 I 2πR center loop: B = µ0 I/2R dQ I= I = −neAvd dt Vs Ns Is Np = , = Vp Np Ip Ns µ χm = −1 µ0 τ =µ×B µ = IA

ξi =Ci (parallel) or Ri (series):

sin(a ± b) = sin a cos b ± cos a sin b

B · dA = 0

F = qv × B; dF = Idl × B µ0 Idl × (r − r ) dB = 4π |r − r |3

1 1 1 = + ξeffective ξ1 ξ2

circ. orbit ˆj ˆ k Ay Az By Bz

î Ax Bx

6.626 × 10−34 J s 1.055 × 10−34 J s 9.11 × 10−31 kg 1.6726 × 10−27 kg 1.6749 × 10−27 kg 1.6605 × 10−27 kg 8.99 × 109 N m2 /C2

h h ¯ = h/2π me mp mn u k

ξi =Ci (series) or Ri (parallel):

Eline =

F = qE dE (at r) =

Planck’s constant Planck’s constant/(2π) electron rest mass proton rest mass neutron rest mass atomic mass unit 1/(4π0 )

a−b a+b sin a + sin b = 2 cos sin 2 2

du u = √ (a2 + u2 )3/2 a2 a2 + u 2

e

du = ln u u 2π

cos2 θ dθ = 0

∂ 2D 1 ∂ 2D = 2 2 ∂x v ∂t2 1 S= (E × B) µ0

√ c = 1/ 0 µ0

c 2 S= = B0 µ0 E0 B0 Erms Brms = = 2µ0 µ0

λ = h/p

1 2

1 un+1 n+1

du 1 = ln(a + bu) a + bu b

1 du = eau a

ln u du = u ln u − u

1 cE02 2 0

un du =

u du 1 = −√ (a2 + u2 )3/2 a2 + u 2 au

ax2 + bx + c = 0 ⇒ √ −b ± b2 − 4ac x= 2a

E×B∝v ∆x∆p > ¯ h ∼

−

2π

sin2 θ dθ = π 0

KE = p2 /(2M ) (plane wave)

p=¯ hk

(¯ h = h/2π)

eiθ = cos θ + i sin θ

E=¯ hω = hf

(de Broglie)

¯ 2 ∂ 2ψ h ∂ψ = i¯ h 2M ∂x2 ∂t

August 16, 2011

Physics 21 Fall, 2011

Solution to HW-2

21-13 Three point charges are arranged on a line. Charge q3 = +5.00 nC and is at the origin. Charge q2 = −3.00 nC and is at x2 = 4.50 cm. Charge q1 is at x1 = 1.00 cm. What is q1 (magnitude and sign) if the net force on q3 is zero? q3

q1

q2

x3=0

x1

x2

We can work this problem without using vectors by thinking it through. Since q2 and q3 have opposite sign, the force on q3 exerted by q2 is attractive (towards the right). If the total force on q3 is to be zero, the force exerted by q1 must be repulsive (toward the left). Thus q1 and q3 must have the same sign, and q1 must be positive. We can find the magnitude of q1 by equating the magnitude of the forces on q3 exerted by q1 and q2 : 1 |q1 q3 | 1 |q2 q3 | = 2 4π0 x1 4π0 x22 Cancelling like terms on both sides of the equation, we find 2 2 x1 1.0 cm |q2 | = (3 nC) = 0.148 nC. |q1 | = x2 4.5 cm We already concluded that q1 was positive. A more general way to solve this problem is to use the vector expressions for the Coulomb force. We want 0 = F1 on 3 + F2 on 3 , where 0=

1 q2 q3 (r3 − r1 ) 1 q1 q3 (r3 − r1 ) + . 4π0 |r3 − r1 |3 4π0 |r3 − r2 |3

and we can evaluate the forces using the locations of the charges. Because q1 is at the origin, r3 = 0. Also, r1 = x1î and r2 = x2î. Substituting for the vectors gives 1 q1 q3 (0 − x1 )î q2 q3 (0 − x2 )î 0= + 4π0 |0 − x1 |3 |x2 |3 q2 x2 ˆ −q3 q1 x1 + = i 4π0 |x1 |2 |x2 |2

21-15 Three point charges are located on the positive x axis of a coordinate system. Charge q1 = 1.0 nC is 2.0 cm from the origin, charge q2 = −4.0 nC is 4.0 cm from the origin and charge q3 = 6.0 nC is located at the origin. What is the net force (magnitude and direction) on charge q1 = 1.0 nC exerted by the other two charges? This problem is very similar to 21-13, and the same diagram applies. Here we need the sum F of F2 on 1 and F3 on 1 , which is q1 q2 (r1 − r2 ) q1 q3 (r1 − r3 ) 1 + F= , 4π0 |r1 − r2 |3 |r1 − r3 |3 where, as before, r3 = 0, r1 = x1î, and r2 = x2î. Then 1 q2 (x1 − x2 )î q3 (x1 − x3 )î F= q1 + 4π0 |x1 − x2 |3 |x1 − x3 |3 q2 (−0.02 m) q3 (0.02 m) ˆ 1 q1 + = i 4π0 (0.02 m)3 (0.02 m)3 Substituting the other numbers leads to −4 nC(−.02 m) 6 nC(.02 m) ˆ 9 F = (9 × 10 )(1 nC) + i (.02 m)3 (.02 m)3 = 2.25 × 10−4 N î

21-11 In an experiment in space, one proton is held fixed and another proton is released from rest a distance d away. What is the initial acceleration of the proton after it is released? From Physics 11 you know that F = ma, or a = F/m. So just find the electrostatic force on one proton due to the other proton, and then divide by the mass. We’ll drop the vector notation and just find the magnitude: a=

e2 1 4π0 mp d2

2 1.602 × 10−19 C = 9 × 10 N m /C (1.67 × 10−26 kg) d2

9

2

2

When you substitute for d, don’t forget to convert to meters. For d = 3 mm = 0.003 m, the result is 2

a = 1.54 × 104 m/s .

Note that the terms in the denominators are lengths and must be positive. The quantity in brackets must be zero, so we obtain 2 3 x2 x1 4.5 1.0 q1 = − q2 = − (−3.0 nC) = 0.148 nC x1 x2 1.0 4.5 This formula agrees with the previous result. Because we were careful with the signs, the formula gives the correct answer for any combination of signs of the three charges and for the two vector components x1 and x2 . (We took x3 = 0.) August 31, 2011

21-46 Two particles having charges q1 = 0.600 nC and q2 = 5.00 nC are separated by a distance of d = 1.60 m. At what point along the line connecting the two charges is the total electric field due to the two charges equal to zero?

q2

q1 0

x

d

Since both charges are positive, it’s easy to keep track of the direction of the electric field. The field at x from q1 points to the right, and the one from q2 points to the left. These two fields must be equal in magnitude for their vector sum to be zero. Therefore q2 1 1 q1 = 2 4π0 x 4π0 (x − d)2 Cancelling the common factor of 1/(4π0 ), we can rewrite the above equation as

(d − x)2 q2 q2 d−x = = ⇒ 2 x q1 x q1

YF 21-50 mod A point charge q1 = −4.00 nC is at the point x = 0.60 m, y = 0.80 m, and a second point charge q2 = +6.00 nC is at the point x = 0.60 m, y = 0. (a,b) Calculate the x and y components of the net electric field at the origin due to these two point charges. (c,d) Calculate the x and y components of the net electric field at the point x = 0.90 m, y = 0.40 m due to these two point charges. Use the vector expression given in class for the field E at r due to a charge Q at point r . Apply this formula to get the field at r due to Q1 ; apply it again to get the field at r due to Q2 , and then add the results (superposition). E (at r) =

Remember, r = field point; r = charge point. y

Q1 = -4.0 nC (0.6, 0.8)

(0.9, 0.4) Q2 = +6.0 nC (0.6, 0.0) x

Solving for x, we find x=

d 1 + q2 /q1

Substituting the specific numbers given above leads to

(a,b) Find E at origin, r = 0î+0ˆj. For Q1 , r = 0.6î+0.8ˆj, so r − r = −0.6î − 0.8ˆj and |r − r | = 1.0 m. For Q2 , r = 0.6î, so r − r = −0.6î and |r − r | = 0.6 m.

x = 0.412 m. Note that instead of taking the square root and solving a linear equation for x, one could also set up a quadratic equation. One must identify the correct root of the quadratic equation, but the result is the same.

1 Q (r − r ) 4π0 |r − r |3

−4 nC(−.6 î − .8 ˆj)m 6 nC(−.6 î)m + (1.0 m)3 (.6 m)3 3.2 nC 1 ˆ 2.4 3.6 = − i + ˆj 4π0 13 (.6)3 13 m2

nC 1 −14.3 î + 3.2 ˆj = 4π0 m2 = −128.3 î + 28.77 ˆj N/C

1 E= 4π0

(c,d) Find E at point r = 0.9 î+0.4 ˆj. For Q1 , r = 0.6î+0.8ˆj, so r−r = 0.3 î−0.4 ˆj and |r − r | = 0.5 m. For Q2 , r = 0.6î, so r − r = 0.3 î + 0.4 ˆj, and |r − r | = 0.5 m.

−4 nC(.3 î − .4 ˆj)m 6 nC(.3 î + .4 ˆj)m + (0.5 m)3 (0.5 m)3 î(−1.2 + 1.8) + ˆj(1.6 + 2.4) nC 1 = 4π0 0.125 m2

nC 1 ˆ 4.8 i + 32 ˆj = 4π m2 0 = 43.2 î + 287.7 ˆj N/C

1 E= 4π0

Physics 21 Fall, 2011

Solution to HW-3

21-96 Positive charge Q is uniformly distributed around a semicircle of radius a. Find the electric field (magnitude and direction) at the center of curvature P .

y

dQ = λds = λadθ

From this result we see the E has a magnitude of Q 2π 2 ǫ0 a2 and points in the −y direction or downward. The x component of the field is zero, as one would expect from symmetry. 21-105 Three charges are placed as shown in the figure. The magnitude of q1 is 2.00 µC, but its sign and the value of the charge q2 are not known. Charge q3 is +4.00 µC, and the net force F on q3 is entirely in the negative x-direction. a) Calculate the magnitude of q2 . b) Determine the magnitude of the net force F on q3 .

dθ x

We start with the equation from the equation sheet that gives the field at the field point r in terms of a charge dQ at the charge point r′ : dE(r) =

1 dQ (r − r′ ) 4πǫ0 |r − r′ |3

The field point r (where we want to know E) is at the origin, so r = 0, and the charge point (where dQ is located) is r′ = a cos θî + a sin θˆj, where θ is the angle above the x-axis, so r − r′ = −a cos θî − a sin θˆj,

and

|r − r′ | = a.

Since we want to add up all the contribution from all dQ, we must relate dQ to dθ so that we can integrate over the angle θ spanned by the semicircle. The arc length ds swept out by an angle dθ is adθ, so the charge dQ on ds is dQ = λds = λa dθ, where λ is the linear charge density (charge per unit length): λ=

Q . πa

With all these substitutions, the original equation becomes 1 ( πa adθ)(−a cos θî − a sin θˆj) dE = 4πǫ0 a3 −Q (cos θî + sin θˆj)dθ. = 4π 2 ǫ0 a2 Q

We can find the total field at the origin by integrating: Z Z π −Q E = dE = (cos θî + sin θˆj) dθ. 2 2 0 4π ǫ0 a We will look at each component of the integral over θ separately. We see that Z π Z π sin θ dθ = 2, cos θ dθ = 0 and 0

0

so we get E=−

Q 4π 2 ǫ0 a2

(0î + 2ˆj) = −

Q 2π 2 ǫ0 a2

ˆj.

(a) We first determine the sign of the charge q1 . We can do this by thinking about which direction the force will be in for the different combinations of signs for charges q1 and q2 . Since there is no y-component of the force on q3 we know that q1 and q2 must have opposite signs. Since the force is directed in the negative x-direction we can infer that q1 must be negative and q2 must be positive. To determine q2 we calculate the the total force F = F1 on 3 + F2 on 3 on q3 . Expressions for F1 on 3 and F2 on 3 follow from the general expression for F2 on 1 on the equation sheet: 1 q3 q2 (r3 − r2 ) 1 q3 q1 (r3 − r1 ) , F2 on 3 . F1 on 3 = 3 4πǫ0 |r3 − r1 | 4πǫ0 |r3 − r2 |3

We need the position vectors of each charge. Let q1 be at the origin. Then the position vectors r1 and r2 of q1 and q2 are trivial. For r3 , we notice that cos θ = 4/5 = x/4 where x is the x-component of the position vector of q3 . Along with this and Pythagorean’s theorem we can find both x and y-components of r3 . The result is r1 = 0, r2 = 5 cm î, r3 = 3.2 cm î + 2.4 cm ˆj

Knowing these position vectors we can write F1 on 3 and F2 on 3 , in terms of the unknown charge q2 : F1 on 3 = −36 N î − 27 N ˆj F2 on 3 = (−24 × 106 N/C) q2 î + (32 × 106 N/C) q2 ˆj

Since the net force on q3 is in the negative x-direction, the sum of the y-components must be 0. From the sum of the y-components we find that the charge q2 = +0.844 µC. The charge is positive, as expected by the reasoning above. (b) Using q2 , we can now determine the net force on q3 by adding the components together. We already know that there is no net y-component, so the net force has only an xcomponent, which is the sum of the x components of F1 on 3 and F2 on 3 . The magnitude of the total force is the absolute value of its x component, |F| = 56.26 N.

September 10, 2011

21-87 A proton with the mass m is projected into a uniform electric field that points vertically upward and has magnitude E. The initial velocity of the proton has a magnitude v0 and is directed at an angle α below the horizontal. (a) Find the maximum distance hmax that the proton descends vertically below its initial elevation. You can ignore gravitational forces. (b) After what horizontal distance d does the proton return to its original elevation? (c) Find the numerical value of hmax if E = 520 N/C, v0 = 5.00 × 104 m/s, and α = 35.0◦ . (d) Find the numerical value of d if E = 520 N/C, v0 = 5.00 × 104 m/s, and α = 35.0◦ . Because the proton is a charged particle (charge e), when it enters a region of uniform electric field, it experiences a constant force according to the relationship F = eE. In previous physics classes, you have studied how a constant force influences the motion of an object. In particular, recall that Newton’s Second Law tells us how force and acceleration are related through the equation F = ma. Combining these two equations allows us to calculate the acceleration: a=

eE . m

Since the electric field is uniform, the acceleration will be constant, and the proton will follow a parabolic trajectory, as shown in the figure.

The acceleration of the proton will be in the same direction as the electric field. Thus, there is zero acceleration along the x direction. This leads to the same type of problem as that studied in projectile motion near the surface of the earth. The only difference is that the acceleration is upward instead of downward. The equations we need to use are: x direction x = x0 + vx0 t vx = vx 0 = constant

However, “height” is a scalar quantity, so we can solve for y1 and take the absolute value. We get an answer of hmax =

mv02 sin2 α . 2eE

(b) In order to calculate x2 , the position when the proton comes back up to its original height, we need to find the time it takes to get there, t2 . Because of the symmetry of the motion, t2 is just twice the time t1 needed to reach the “peak”. Using vy = vy0 + at, we substitute in vy = 0 at t1 , use our expression for vy0 , and solve for t1 . Doubling this result gives us t2 =

2v0 sin α . a

Substitute this expression into x = x0 + vx0 t, use our expression for a, and set x0 = 0 to get x2 =

2mv02 sin α cos α . eE

This expression is also the answer for d, the distance travelled in the x direction, since d = x2 − x0 = x2 − 0. (c) and (d) Substituting in the given values, along with the fundamental constants e = 1.602 × 10−19 C and m = 1.673 × 10−27 kg (see the equation sheet), yields hmax = 8.26 × 10−3 m

and d = 4.72 × 10−2 m.

21-99 Two 1.20 m nonconducting wires meet at a right angle. One segment carries 2.00 µC of charge distributed uniformly along its length, and the other carries −2.00 µC distributed uniformly along it, as shown in the figure. Find (a) the magnitude and (b) the direction of the electric field these wires produce at point P , which is 60.0 cm from each wire. If an electron is released at P , what is (c) the magnitude and (d) the direction of the net force that these wires exert on it?

y direction y = y0 + vy 0 t + 12 at2 vy = vy 0 + at 2 + 2a∆y vy2 = vy0

Before we go any further, we must resolve the intial velocity vector into x and y components: v0 = vx0 î + vy0 ˆj, where vx0 = v0 cos α

and vy0 = −v0 sin α.

(a) When the proton reaches its maximum “height”, the y component of the velocity reaches 0 (see v1 in the figure). 2 Substituting into vy2 = vy0 + 2a∆y yields 0 = v02 sin2 α + 2ay1 , where y1 is the position at maximum “height”. By the choice of coordinate system, this will be a negative position.

(a) In class we worked out the vector electric field due to a finite line of charge at a field point located directly outward from the midpoint of the line. The magnitude of the field is E=

1 λ a √ , 2πǫ0 d d2 + a2

where d is the distance of the point from the line of charge, and a is half the length of the line. In this exercise, the two wires have the same length, the same magnitude of charge,

and are the same distance from the point P . Thus, we can adapt the formula above to the charge and spatial orientation of each wire by putting in the correct direction for each field. Recalling that λ = Q/L and a = L/2, we find E=

1 1 Q p 4πǫ0 d d2 + (L/2)2

= (9 × 109 Nm2 /C2 ) = 35355N/C.

2 × 10−6 C 1 p 0.6m (0.6 m)2 + (0.6 m)2

The field E− due to the negatively charged wire is directed to the left, while the field E+ due to the positively charged wire is straight down. We can use superposition to find the net electric field Enet Enet

= E− + E+ = −35355 N/C î − 35355 N/C ˆj q 2 + E 2 = 5.00 × 104 N/C = E− +

(b) Since the field vectors are perpendicular and of equal magnitude, the net electric field will be 45◦ from either vector, and 135◦ counterclockwise from the +y axis. (c) The force acting on a charge q is F = qE, so at P the magnitude of the force on the electron is Fnet = eEnet = (1.60 × 10−19 C)(5.00 × 104 N/C) = 8.00 × 10−15 N.

(d) Since the electron has a negative charge, the force will be directed opposite (180◦ ) from the electric field. Counterclockwise from the +y axis, θ = 315◦ . 21-104 A thin disk with a circular hole at its center, called an annulus, has inner radius R1 and outer radius R2 . The disk has a uniform positive surface charge density σ on its surface. (a) Determine the total electric charge on the annulus. (b) The annulus lies in the yz -plane, with its center at the origin. For an arbitrary point on the x -axis (the axis of the annulus), find the magnitude of the electric field E. Consider points above the annulus in the figure. (c) Find the direction of the electric field E. Consider points above the annulus in the figure.

surface area. For an annulus, A = πR22 − πR12 . Mastering Physics is expecting a symbolic expression, so simply enter ¢ ¡ σπ R22 − R12

(b) This problem is similar to Example 21.12 in the textbook. Our target is the electric field along a symmetry axis of a continuous charge distribution. We can represent the charge distribution as a collection of concentric rings of charge dQ. From lecture, we know the field of a single ring on its axis of symmetry, so all we have to do is add the contributions of the rings. For a single ring, Ering = î

Qx 1 4πǫ0 (x2 + a2 )3/2

where Q is the total charge on the ring, x is the distance along the axis from the ring to the point at which we are finding the field, and a is the radius of the ring. We will now assume an infinitesimally thin ring, with charge dQ on it, of radius r′ , and thickness dr′ . It will have a contribution to the electric field 1 dQ x dE = î 4πǫ0 (x2 + r′2 )3/2 Before we can integrate this, we need to come up with an expression that tells us how much charge is on our infinitesimally thin ring. We will approximate the area of the ring as circumference × thickness (this approximation works because the ring is infinitesimally thin). So, dQ = σdA = σ2πr′ dr′ . We now have r′ dr′ σx dE = î 2ǫ0 (x2 + r′2 )3/2 To get the net electric field, we integrate this expression from r′ = R1 to r′ = R2 . ¸R2 · Z σx R2 r′ dr′ î σx − √ 1 E = î = 2ǫ0 R1 (x2 + r′2 )3/2 2ǫ0 x2 + r′2 R1   1 σ  1  q −q = î 2ǫ0 2 2 1 + (R1 /x) 1 + (R2 /x) Note that the integral used here is on the table of integrals on the equation sheet. Mastering Physics is only asking for the magnitude of the electric field here, so enter   σ  1 1  q −q 2ǫ0 2 2 1 + (R1 /x) 1 + (R2 /x) (c) Here we select the direction of the electric field that was found in part (b), the positive x -direction.

(a) Given a uniform positive surface charge density σ, the total electric charge on any surface is σA, where A is the

Physics 21 Fall, 2011

Solution to HW-4

22-7 The electric field due to an infinite line of charge is perpendicular to the line and has magnitude E = λ/2π0 r. Consider an imaginary cylinder with a radius of r = 0.200 m and length l = 0.465 m that has an infinite line of positive charge running along its axis. The charge per unit length on the line is λ = 7.15µC/ m. (a) What is the electric flux through the cylinder due to this infinite line of charge? (b) What is the flux through the cylinder if its radius is increased to r = 0.515 m? c) What is the flux through the cylinder if its length is increased to l = 0.760 m?

22-10 A point charge q1 = 4.00 nC is located on the x-axis at x = 2.00 m, and a second point charge q2 = −6.00 nC is on the y-axis at y = 1.00 m. What is the total electric flux due to these two point charges through a spherical surface centered at the origin and with radius (a) 0.500 m, (b) 1.50 m, (c) 2.50 m?

q2 Sa

q1 Sb Sc

The problem asks for total electric flux through a surface. Gauss’s Law helps here. The Equation Sheet gives Gauss’s Law in the form: Q E · dA = , 0

(a) Because the electric field lines are perpendicular to the line of charge, they will also be perpendicular to the surface of the curved or “barrel” part of the cylinder, and parallel to the ends of the cylinder. Because the line of charge is concentric with the cylinder, the field strength will also be constant along the barrel part of the cylinder. Then the electric flux can just be calculated by: Φ = EAbarrel =

λ 2πrl = 375, 700 Nm2 /C 2π0 r

Note that the r’s cancel; the flux is independent of the radius of the cylinder. We can also use Gauss’ law to do the calculation and obtain the same expression. Φ=

where the left hand side is the total electric flux ΦE through a surface S, and Q is the charge enclosed by S. (a) Surface Sa , whose radius is 0.500 m, encloses no charge. The total electric flux is zero (in any units). (b) Surface Sb encloses q2 but not q1 , so Q = q2 : ΦE =

q2 −6.00 × 10−9 C = −678 NC−1 m2 . = 0 8.85 × 10−12 C2 Nm− 2

Note the units: NC−1 is electric field and m2 is area. (c) Surface Sc encloses both charges, so Q = q1 + q2 : q1 + q2 4.00 × 10−9 − 6.00 × 10−9 = 0 8.85 × 10−12 = −226 NC−1 m2 .

ΦE =

λl Qenc = = 375, 700 Nm2 /C 0 0

(b) We already found in (a) that the flux is independent of r; Gauss’ law also tells us that the flux through the cylinder only depends on the charge enclosed. Because increasing the radius does not change how much charge is enclosed, the answer for part (b) should be the same as the answer for part (a). Φ = 375, 700 Nm2 /C (c) If the length of the cylinder is increased, the new flux must be calculated because now more charge on the line is enclosed. Φ=

λl Qenclosed = = 614, 000 Nm2 /C 0 0

September 12, 2011

22-20 The electric field 0.400m from a very long uniform line of charge is 840 N/C. How much charge is contatined in a 2.00-cm section of the line? The equation for the electric field of an infinite wire is E=

1 λ , 2π0 r

22-24 A point charge of −2.11 µC is located in the center of a spherical cavity of radius rcav = 6.50 cm inside an insulating spherical charged solid. The charge density in the solid is ρ = 7.36 × 10−4 C/m3 . Calculate (a) the magnitude and (b) the direction of the electric field inside the solid at a distance r = 9.48 cm from the center of the cavity.

where r is the perpendicular distance to the wire, and λ is charge per unit length. Solving for λ gives λ = 2π0 rE = 2π × 8.8541 × 10−12 × 0.4 × 840 = 1.86 × 10−8

C m

If we muliply λ (in C/m) by the length (in m) we get charge: λ × 0.02 = 3.74 × 10−10 C 22-26 A conductor with an inner cavity, like that shown below, carries a total charge of Qcond = +4.80 nC. An additional charge within the cavity, insulated from the conductor, is q = −6.20 nC. (a) How much charge is on the inner surface of the conductor? (b) How much charge is on the outer surface of the conductor?

(a) The Gaussian surface G shown includes the inner surface of the conductor. The electric field must be zero inside the conductor, making G E · dA = 0, and thus the total electric charge Qencl enclosed by G must be zero. The charges contained within the surface are the charge in the cavity q and the charge on the conductor’s inner surface qinner , so Qencl = 0 = qinner + q

⇒

qinner = −q = 6.20 nC.

(b) The total charge on the conductor Qcond is the sum of the charge on the inner surface qinner and the charge on the outer surface qouter . Hence Qcond = 4.8 nC = qinner + qouter qouter = 4.8 nC − qinner = 4.8 nC − 6.20 nC = −1.4 nC

(a) To find the electric field we can use Gauss’ Law, Qencl E · dA = , 0 S where we choose the Gaussian surface S to be a sphere of radius r = 9.48 cm centered at the point charge. At every point on S the vector dA and E are perpendicular to S and are therefore parallel (or antiparallel – we’ll provisionally assume the former). By symmetry, the electric field has the same magnitude E everywhere on the surface, so we can evaluate the surface integral in terms of the unknown E, Qencl E · dA = E dA = 4πr2 E = . 0 S S All that remains is to determine the charge enclosed by the Gaussian surface, which consists of the point charge qpoint and the charge Qinsul on the insulator, which we can calculate as ρV . (Note that the charge on an insulator can be evenly distributed throughout its volume with density ρ.) The volume V of the insulator inside S is the volume of S less the volume of the cavity, 4 4 3 4 3 3 πr − πrcav = π(r3 − rcav ) 3 3 3 4 = π (.0948 m)3 − (.0650 m)3 = .00242 m3 . 3

V =

Thus the total charge enclosed is given by Qencl = qpoint + Qinsul = qpoint + ρVencl = − 2.11×10−6 C+(.000736 C/m3 )(.00242 m3 ) = − 3.3 × 10−7 C. Now we can use the relation between E and Qencl : E= =

Qencl 4πr2 0 4π(9.48 ×

10−2

−3.3 × 10−7 C m)2 (8.854 × 10−12 C2 /(Nm2 ))

= − 3.3 × 10−5 N/C. (b) Since the magnitude E must be positive, the minus sign we obtained for E above tells us that E and dA must be antiparallel, that is, E points into the surface.

Physics 21 Fall, 2011

Solution to HW-5

22-4 A cube has sides of length L = 0.330m. It is placed with one corner at the origin as shown in the figure. The electric field is not uniform but is given by

For surface S5 , the normal direction is in the +x direction, so we integrate the î component over the surface. Φ5 = −

Z

0

L

Z

L

4.41x dy dz = −4.41xL2

0

Evaluating this at x = L, gives the flux:

ˆ E = [−4.41 N/(C m)] x î + [3.29 N/(C m] z k.

Φ5 = −4.41L3 = 0.158 N m2 /C.

(a) Find the electric flux through each of the six cube faces S1 , S2 , S3 , S4 , S5 , S6 . (b) Find the total electric charge inside the cube.

Surface S6 is similar, but the surface is located at x = 0 so the flux evaluates to Φ6 = 0. (b) To find the total electric charge inside the cube, we can apply Gauss’ Law, which tells us that the total flux through this cube is equal to the charge enclosed, divided by ǫ0 . Φ1 + Φ2 + Φ3 + Φ4 + Φ5 + Φ6 =

Qencl ǫ0

Solving for Qencl and substituting the numbers gives Qencl = (0 + 0.118 + 0 + 0 − 0.158 + 0)ǫ0 = −3.54 · 10−13 C.

R (a) The electric flux is defined as Φ = E · dA. Because dA is always perpendicular to the surface, we only need the component of the field that is normal to the surface of interest. (The dot product of dA with the parallel component of the field will be zero.) For each of the six surfaces, it will work out that the component normal to the surface is constant, so the surface integral will just be the constant value of the normal component times the surface area L2 . For surface S1 , there is no component of the electric field normal to the surface (in the y direction ˆj), so the electric flux Φ1 = 0. For surface S2 , the component of the electric field normal ˆ to the surface is in the +z direction, so take only the k component of E. The integral over the surface is Z

L

0

Z

L

3.29z dx dy = 3.29zL2 .

0

To evaluate the electric field at the surface we set z = L, and then the flux is Φ2 = 3.29L3 = 0.118 N m2 /C Surface S3 is similar to S1 ; there is no electric field component in the normal direction ˆj, so the flux Φ3 = 0. For surface S4 , the normal direction is in the −z direction, ˆ component. The so we need to take the negative of the k flux is that component integrated over the surface. Φ4 = −

Z

0

L

Z

22-26 A conductor with an inner cavity, like that shown below, carries a total charge of qtot = +5.80 nC. The charge within the cavity, insulated from the conductor, is qcenter = −7.10 nC. (a) How much charge is on the inner surface of the conductor? (b) How much charge is on the outer surface of the conductor? qouter qinner

qtot = +5.80 nC Gaussian Surface

qcenter -7.10 nC

(a) The insulated point in the center of the cavity will induce an opposite charge on the inner surface of the conductor. The induced charge will have the same magnitude as the center charge, qinner = +7.10 nC. We can prove this by placing a Gaussian surface within the conductor that encompasses the entire cavity and point charge. Inside the conductor E = 0, and therefor Qencl = qcenter + qinner = 0. (b) The total charge qtot on the conductor is a constant and does not change, but we must find how it is divided between the inner and the outer surface. qtot = qinner + qouter =⇒ qouter = qtot − qinner qouter = 5.80 nC − 7.10 nC = −1.30 nC

L

3.29z dx dy = −3.29zL2

0

But evaluating the electric field at z = 0 results in Φ4 = 0.

September 15, 2011

23-8 Three equal 1.20µC point charges are placed at the corners of an equilateral triangle whose sides are 0.500 m long. What is the potential energy of the system? (Take as zero the potential energy of the three charges when they are infinitely far apart.) The total potential energy is obtained by summing up the potential energy of interaction between each pair of charges (superposition). If we label the three charges 1, 2, and 3 then U = U12 + U13 + U23

q1 q3 q2 q3 1 q1 q2 ( + + ) = 4πǫ0 r12 r13 r23

Since all three charges are identical and the separation of any pair is the same, then U=

3(1.2 × 10−6 µC)2 1 3q 2 = (9 × 109 ) = 0.078 J 4πǫ0 r 0.500m

23-5 A small metal sphere, carrying a net charge of q1 = −2.90 µC, is held in a stationary position by insulating supports. A second small metal sphere, with a net charge of q2 = −7.80 µC and mass 1.80 g, is projected toward q1 . When the two spheres are ri = 0.800 m apart, q2 is moving toward q1 with speed vi = 22.0 m/s. Assume that the two spheres can be treated as point charges. You can ignore the force of gravity. (a) What is the speed of q2 when the spheres are rf = 0.430 m apart? (b) How close does q2 get to q1 ?

For part (a) we are given rf , and we need to solve Eq. (1) for vf . The result is vf = 15.5 m/s. For part (b), we know that when q2 is closest to q1 , the KE reaches its minimum value of zero, so we must set vf = 0 in Eq. (1) and solve for rf . The result is rf = 0.295 m. 23-79 Electric charge is distributed uniformly along a thin rod of length a, with total charge Q. Take the potential to be zero at infinity. (a) Find the potential at the point P , a distance x to the right of the rod. (b) Find the potential at the point R, a distance y above the right-hand end of the rod. (c) In part a, what does your result reduce to as x becomes much larger than a? (d) In part b, what does your result reduce to as y becomes much larger than a?

We will express the potential dV at points P and R from charge dQ at the charge point r′ = x′ î, and then we integrate over the length of the charged rod (from x′ = −a to x′ = 0). The general formula is dQ 1 . dV = 4πǫ0 |r − r′ | The charge dQ can be related to the length dx′ using the linear charge density: Q dQ = λdx′ = dx′ . a ′ ˆ (a) For point P , r = x i and r − r = (x − x′ ) î. Hence

For this problem we will use the conservation of energy. We first notice that the two charges are both negative, so the Coulomb force is repulsive. Particle 2 must therefore start with positive kinetic energy KE, and as it approaches particle 1, KE will diminish and the potential energy U will increase by the same amount. We know KE = 21 mv 2

and

U=

1 q1 q2 , 4πǫ0 |r1 − r2 |

1 q1 q2 = 0.689841 J. 4πǫ0 ri

Since energy is conserved, Etot has the same value at any other value of rf : Etot = 12 mvf2 +

1 q1 q2 . 4πǫ0 rf

By equating the two expressions for Etot , we see 2 1 2 mvf

+

1 q1 q2 = 0.689841 J. 4πǫ0 rf

V =

Z

0

−a

dV =

1 Q 4πǫ0 a

Z

0

−a

dx′ 1 Q x+a = ln x − x′ 4πǫ0 a x

(b) For point R, r = y ˆj and r − r′ = −x′ î + y ˆj. Hence dx′ 1 Q p 4πǫ0 a (−x′ )2 + y 2 Z Q dx′ y 1 Q 0 p = ln p V= 2 4πǫ0 a −a (x′ )2 + y 2 4πǫ0 a a + y2 − a

dV =

where |r1 − r2 | is the distance between the charges. We can evaluate the total energy Etot by substituting the given initial values ri and vi : Etot = 21 mvi2 +

1 Q dx′ 4πǫ0 a (x − x′ )

dV =

(1)

Another expression for the argument of the log follows from p y a2 + y 2 + a p = y a2 + y 2 − a (c,d) For x or y large compared to a, we can consider the whole charge Q on the line to be a point charge at the origin. The potential is then 1/(4πǫ) times Q/x for c or times Q/y for d. Mathematically the result follows if we use the approximation ln(1 + δ) ≈ δ for δ ≪ 1. The key steps are x+a a a ln = ln(1 + ) ≈ x px px ¶ µ a a2 + y 2 + a y2 + a a ≈ . ≈ ln ≈ ln 1 + ln y y y y

Physics 21 Fall, 2011

Solution to HW-6

23-9 A point charge q1 = 4.10 nC is placed at the origin, and a second point charge q2 = −2.9 nC is placed on the x-axis at x = +21.0 cm. A third point charge q3 = 2.1 nC is to be placed on the x-axis between q1 and q2 . (Take as zero the potential energy of the three charges when they are infinitely far apart.) a) What is the potential energy of the system of the three charges if q3 is placed at x = +11.0 cm? b) Where should q3 be placed to make the potential energy of the system equal to zero? (a) The general formula for the potential energy between any two point charges labeled 1 and 2 is U12 =

1 q1 q2 , 4πǫ0 |r1 − r2 |

where |r1 − r2 | is the distance between the two charges. To find the total potential energy of several charges, we must find the potential energy due to each pair of point charges. So the total potential energy here is Utotal = U12 + U13 + U23 ¶ µ q1 q3 q2 q3 q1 q2 1 + + = 4πǫ0 |0 − .21 m| |0 − .11 m| |.21 m − .11 m| = −3.53 × 10−7 J

(b) We can say that q3 is placed at some point x between the other two charges and determine which value of x gives the system a total potential energy of zero. Incorporating this into the expression for the total potential energy gives: ¶ µ q1 q3 q2 q3 q1 q2 1 =0 + + Utotal = 4πǫ0 .21 m x (.21 m − x) We can save ourselves some trouble by eliminating several overall factors that won’t affect the solution x. Since all the terms have exactly the same units, we can drop the conversion factors to SI units and write distances in cm and charges in nC. The result is (4.1)(−2.9) (4.1)(2.1) (−2.9)(2.1) + + =0 21 x (21 − x) Multiply through by x(21 − x) to get a quadratic equation:

23-31 (a) An electron is to be accelerated from a velocity of 2.50 × 106 m/s to a velocity of 8.00 × 106 m/s. Through what potential difference must the electron pass to accomplish this? (b) Through what potential difference must the electron pass if it is to be slowed from 8.00 × 106 m/s to a halt?

This problem is a conservation of energy problem. The only forms of energy involved are electrical potential energy (U ) and kinetic energy (K). Note that “potential difference” means the “change in voltage” (V2 −V1 ) from the initial value V1 to the final value V2 , and that the relationship U = qV is not on your equation sheet. (a) K1 + U1 2 1 2 mv1

= K2 + U2

+ qV1 = 21 mv22 + qV2

q(V2 − V1 ) = − 21 m(v22 − v12 ) The final equation shows that the change in potential energy is the negative of the change in kinetic energy, as expected. Dividing by the charge gives V2 − V1 = −

m 2 (v − v12 ). 2q 2

Using the velocities given and remembering that q = −e for an electron gives us the answer V2 − V1 = 164 V. Think carefully about the sign. A positively charged particle would slow down when the potential goes up (like a mass going uphill), but an electron behaves in the opposite way. (b) With the same equation we found for part (a), we can recalculate for the new velocities that V2 − V1 = −182 V. In this case, the particle slows down, so a positively charged particle would go through a positive change in potential, but again the electron does the opposite. Note that on Mastering Physics, the question asks for the value of V1 − V2 , not V2 − V1 . So, the numerical answers you should give are −164 V and 182 V.

0.5662x2 − 26.59x + 180.81 = 0 Using the quadratic formula we find x = 38.72 or 8.25. Remember that these values are in centimeters. Since the first solution is not between the two other point charges, we can ignore it. So q3 must be placed in between the two charges at x = 8.25 cm to give the system a total potential energy of zero.

September 16, 2011

23-36 A very long insulating cylinder of charge of radius 3.00 cm carries a uniform linear density of 15.0 nC/m. If you put one probe of a voltmeter at the surface, how far from the surface must the other probe be placed so that the voltmeter reads 175 V? The symmetry of such a uniformly charged cylinder is the same as the symmetry of a uniform line of charge. In fact, outside of the cylinder, the electric field is the same as though all the charges were concentrated on a line along its axis. (We can see this by applying Gauss’s law to a Gaussian surface like that used for the uniform line of charge.) This electric 1 λ where λ field is listed on the equation sheet as Eline = 2πǫ 0 r is the linear charge density and r is the distance away from the line. The electric field lines are directed radially outward from the surface of the cylinder. Now that we know what the electric field looks like, we can calculate the potential difference (the quantity measured by the voltmeter) by evaluating the line integral of the field between the probes of the voltmeter. We will evaluate Z f E · dl Vf − Vi = −

From the second equation, Ey = 0 when x = −C/A. Then (from the first equation) Ex = 0 when −Ay + 2B(−C/A) = 0

Be careful with the signs. The argument of the logarithm is greater than one, so the logarithm is positive. Thus ∆V is negative. That makes sense; the potential decreases as one goes further away from the positively charged cylinder. Now, we solve for d and evaluate. µ ¶ −2πǫ0 ∆V d = ln 1 + λ r0 ¶ µ d −2πǫ0 ∆V =1+ exp λ r0 · µ ¶ ¸ −2πǫ0 ∆V − 1 = 2.74 cm d = r0 exp λ Again, we used ∆V = −175 V.

23-47 In a certain region, the electric potential is V (x, y, z) = Axy −Bx2 +Cy, where A, B, and C are positive constants. Find Ex , Ey , and Ez . Where is the electric field zero (multiple choice)? ∂V = −Ay + 2Bx ∂x ∂V Ey = − = −Ax − C ∂y ∂V Ez = − =0 ∂z Ex = −

y = −2BC/A2 .

That analysis gives x and y; any value of z is OK. 23-66 A disk with radius R has a uniform charge density σ. (a) By regarding the disk as a series of thin concentric rings, calculate the electric potential V at a point on the disk’s axis a distance x from the center of the disk. Assume that the potential is zero at infinity. (Hint: Use the result that the potential at a point on the ring axis at a distance x from the center of the ring is V =

Q 1 √ 4πǫ0 x2 + a2

where Q is the charge of the ring.) (b) Find Ex = −∂V /∂x. (a) Using the hint, we can modify the result for the finite ring to determine the potential dV due to a thin ring of radius r and charge dQ. We have

i

along a path radially outward, beginning at the surface of the cylinder. This will make E and dl parallel, which simplifies the dot product in the integral. Let’s call the radius of the cylinder r0 and call the distance away from the cylinder’s surface that we are looking for d. Z r=r0 +d Z f 1 λ dr Edl = − ∆V = − 2πǫ 0 r r=r0 i Z r=r0 +d λ dr λ = − =− [ln (r0 + d) − ln r0 ] 2πǫ0 r=r0 r 2πǫ0 ¶ µ d λ ln 1 + = − 2πǫ0 r0

⇒

dV =

1 dQ √ . 4πǫ0 x2 + r2

Because electric potential is a scalar quantity, we can integrate dV without keeping track of vectors. First, we must determine dQ in terms of known quantities. It is equal to the surface charge density times the surface area dA of the thin ring, dQ = σdA = σ2πrdr, where dA is the product of the length (circumference) 2πr and width dr of the ring of radius r. Now it is possible to do the integration Z Z R Z R 1 2πσr dr r dr σ √ √ V = dV = = 2 2 4πǫ0 0 2ǫ x +r x2 + r2 0 0 The integral is on the equation sheet: Z p udu √ = a2 + u2 . 2 2 a +u Evaluating it at the limits results in i σ hp 2 x + R2 − x . V = 2ǫ0 (b) The x component of the electric field is equal to −∂V /∂x: · ¸ ¢ 1 ∂V σ 1¡ 2 2 −2 − =− 2x − 1 x +R ∂x 2ǫ0 2 ¸ · σ x . = 1− √ 2 2ǫ0 x + R2 Notice that as R → ∞, this result reduces to the field of an infinite sheet.

Physics 21 Fall, 2011

Solution to HW-7

24-25 A 6.60 µF, parallel-plate, air capacitor has a plate separation of 3.00 mm and is charged to a potential difference of 300 V. Calculate the energy density in the region between the plates. Energy density is calculated using the formula u = 21 ǫ0 E 2 . In this problem, we know the voltage on the capacitor, not the electric field strength between the plates. But the voltage and electric field strength are related. Because the electric field between the plates is uniform and directed straight across the gap between the plates, it is easy to show that V = Ed where d is the separation between the plates. Doing the substitutions and evaluating for the given values, we get µ ¶2 µ ¶2 ¶µ 2 V 300 V −12 C 1 1 = 2 8.85 × 10 u = 2 ǫ0 d N · m2 .003 m = 4.43 × 10

−2

(a) The reciprocal of the equivalent capacitance for capacitors in series is the sum of the reciprocals of each capacitor. The equivalent capacitance for capacitors in parallel is the sum of each capacitor. Using these rules we can break down the circuit by combining each capacitor until we are left with a single equivalent capacitance. C3 and C4 are in series so we can find the equivalent capacitance: 1 1 C3 C4 1 = + =⇒ C3,4 = . C3,4 C3 C4 C3 + C4 Substituting, we find C3,4 = 2.4 µF. Next we combine this with C2 . The capacitors are in parallel so we find: C2,3,4 = C2 + C3,4 = 7.2 µF. Finally, we combine C1 , C2,3,4 , and C5 which are in series. 1 1 1 1 = + + . Ceq C1 C2,3,4 C5 We find Ceq = 2.63 µF.

3

J/m .

24-38 A parallel-plate capacitor has capacitance C0 = 5.00 pF when there is air between the plates. The separation between the plates is 1.50 mm. (a) what is the maximum magnitude of charge Q that can be placed on each plate if the electric field in the region between the plates is not to exceed 3.00 × 104 V/m? (b) A dielectric with K = 2.70 is inserted between the plates of the capacitor, completely filling the volume between the plates. Now what is the maximum magnitude of charge on each plate if the electric field between the plates is not to exceed 3.00 × 104 V/m? (a) Without the dielectric, Q = C0 V and V = Ed. Therefore, the charge for the given values of C0 , E, and d are µ ¶ ¡ ¢ 4 V −12 Q = C0 Ed = 5.00×10 F 3.00×10 (.0015 m) m = 2.25 × 10−10 C = 225 pC

(b) With the dielectric, C increases to KC0 , and V = Ed still holds. The charge is K times the previous value: ¡ ¢ Q = KC0 Ed = 2.7 × 2.25 × 10−10 C = 6.08 × 10−10 C = 608 pC

24-59 In the figure, C1 = C5 = 8.3 µF and C2 = C3 = C4 = 4.8 µF. The applied potential is Vab = 230 V. (a) What is the equivalent capacitance of the network between points a and b? (b) Calculate the charge on each capacitor and the potential difference across each capacitor.

(b) We know that for capacitors in series, the charge on the equivalent capacitor is the same as the charge on each individual capacitor. Also, for capacitors in parallel, the potential difference across the equivalent capacitor is the same as the potential difference across each individual capacitor. Keeping these rules in mind, we will determine the charge and potential difference across each capacitor by rebuilding circuit in the opposite way that we broke it down in part (a), calculating the charge and potential at each point. We can determine the total charge on Ceq since we know the potential across a and b. Qtot = Ceq Vab = 6.1 × 10−4 C = 610 µC. Ceq is made up of C1 , C2,3,4 , and C5 , which are in series, so this charge is the charge on each of these three capacitors, so we know Q1 = Q5 = 610 µC. We can find the potential difference by applying V = Q/C to find V1 = V5 = 73 V.

We can determine the potential across C2,3,4 the same way to find V2,3,4 = 84 V. Again, C2,3,4 is made up of C2 and C3,4 , which are in parallel, so this is the potential difference across both capacitors. So we know V2 = V3,4 = 84 V, and applying Q = CV gives Q2 = 400 µC and Q3,4 = 200 µC.

C3,4 is made up of C3 and C4 in series so we know Q3,4 = Q3 = Q4 = 200 µC. By applying V = Q/C we find V3 = V4 = 42 V. Capacitor 1 2 3 4 5

Charge 610 µC 400 µC 200 µC 200 µC 610 µC

Potential Difference 73 V 84 V 42 V 42 V 73 V

September 21, 2011

24-61 Three capacitors having capacitances of 8.0 µF, 8.3 µF, and 4.1 µF are connected in series across a 40 V potential difference. (a) What is the charge on the 4.1 µF capacitor? (b) What is the total energy stored in all three capacitors? (c) The capacitors are disconnected from the potential difference without allowing them to discharge. They are then reconnected in parallel with each other, with the positively charged plates connected together. What is the voltage across each capacitor in the parallel combination? (d) What is the total energy now stored in the capacitors?

The parallel combination is equivalent to a single effective capacitance Ceff = C1 + C2 + C2 with the same total charge 3Q. The voltage V = Va − Vb is then given by ¡ ¢ 3 8.2 × 10−5 C 3Q Qtot = = = 12 V V = Ceff C1 + C2 + C3 (8.0 + 8.3 + 4.1) µF One could determine the charges on each capacitor with this result, using Qi = Ci V . (d) As in part (b), the total energy stored in all the capacitors can be found by adding up the energies on each capacitor. This time use the form of the equation that involves the voltage and the capacitance: Ucap = 21 C1 V 2 + 12 C2 V 2 + 21 C3 V 2 =

1 2

(C1 +C2 +C3 ) V 2

= 21 Ceff V 2 = 21 (2.04 × 10−5 F)(12 V)2 = 1.5 mJ.

(a) In a series connection the charges on the plates is as shown. We can determine the effective capacitance of the three capacitors and use this to determine the charge Q. The effective capacitance is found using 1 1 1 1 1 1 1 = + + = + + Ceff C1 C2 C3 8.0 µF 8.3 µF 4.1 µF Ceff = 2.04 µF Then

24-65 A parallel-plate capacitor with only air between the plates is charged by connecting it to a battery. The capacitor is then disconnected from the battery, without any of the charge leaving the plates. (a) A voltmeter reads 41.0 V when placed across the capacitor. When a dielectric is inserted between the plates, completely filling the space, the voltmeter reads 11.5 V. What is the dielectric constant of this material? (b) What will the voltmeter read if the dielectric is now pulled partway out so it fills only one third of the space between the plates? (a) With no dielectric, Q = C0 V0 , where V0 has been measured. With the dielectric, the charge does not change, but a new V is measured, so Q = KC0 V . Equating both expressions for Q, C0 V0 = KC0 V

Q = Ceff V = (2.04 µF) × (40 V) = 8.2 × 10−5 C.

K=

V0 41 = = 3.57 V V 11.5

(b) This situation is essentialy two capacitors in parallel:

(b) The total energy stored in all the capacitors can be found by adding up the energies on each capacitor. Use the form of the equation that involves the charge and the capacitance: µ ¶ 1 Q2 Q2 Q2 1 1 Ucap = 12 + 12 + 21 = 21 Q2 + + C1 C2 C3 C1 C2 C3 ¡ ¢2 −5 8.2 × 10 C Q2 = 12 = 12 = 1.6 mJ Ceff 2.04 × 10−6 F Note one gets the same answer using the form Ucap = 21 CV 2 , with Ceff and V = 40 V. (c) When the capacitors are disconnected from the serial circuit, the charge on each one remains the same (+Q). When they are reconnected in parallel, the total charge (3Q) is free to move around on the three positive plates shown in the diagram. After the charges equilibrate, the potential difference Va − Vb must be the same across every one of the capacitors.

⇒

(2/3) C0

(1/3) KC0

Ceff = 23 C0 + 31 KC0 Therefore the measured voltage will be V =

Q = Ceff

2 3 C0

C0 V0 = + 31 KC0

2 3

V0 3V0 = 22.1 V = 1 K +2 + 3K

Physics 21 Fall, 2011

Solution to HW-8

25-44 If a “75 W” bulb (75 W are dissipated when connected across 120V) is connected across a 220 V potential difference (as is used in Europe), how much power does it dissipate? (a) The power dissipation P and potential difference V across the bulb are variables, but the resistance R across the light bulb is a physical constant whether you are in America or Europe. From the equation sheet we know that P = IV and V = IR. We can combine these formulas to eliminate the current I, which we don’t know, and get a relation between power, voltage, and resistance: P = IV =

V2 V2 V V = ⇒R= . R R P

Using the US values P = 75 W and V = 120 V, we find R=

(120 V)2 = 192 Ω. 75 W

We can then use this resistance to solve for the power dissipated when the bulb is connected to the new potential difference: P =

V2 (220 V)2 = = 252 W. R 192 Ω

25-46 A battery-powered global positioning system (GPS) receiver operating on a voltage of 9.1 V draws a current of 0.20 A. a) How much electrical energy does it consume during a time of 2.0 h?

(a) We can model the power supply as a battery with V = 14 kV in series with an internal resistor of Rint . When the person holds the terminals of the power supply (don’t do this at home, kids), he completes the circuit shown in the diagram. The total resistance of resistors in series is the sum Rtot = Rint + Rperson , and the current going through them will be the same as the current going through the equivalent Rtot . This current is I=

V 14 kV V = = = 1.167 A. Rtot Rint + Rperson 2 kΩ + 10 kΩ

(b) The power dissipated in a device with current running through it is given by: P = IV. We already found the current going through the person in part (a). However the potential V here is not the same potential as the battery. It is the potential across just the person’s body, which we can find using V = IRperson : 2

P = IV = I 2 Rperson = (1.167 A) 10 kΩ = 13.6 kW. (c) To make the battery safe we need to increase the internal safe resistance to Rint . From the second equation in part (a) we can see that V . Isafe = safe Rint + Rperson safe Solving this equation for Rint , we have safe Rint =

14 kV V − Rperson = − 10 kΩ = 13.99 MΩ. 0.001 A 0.001 A

(a) We can calculate the energy use over a period of time if we know the rate that the device uses electrical energy (power). The power can be determined from: P = IV = 9.1 V × 0.2 A = 1.82 J/s = 1.82 W The total energy used over a time of two hours is: Etot = P t = 1.82 J/s × 7200 s = 1.31 × 104 J 25-70 A person with body resistance between his hands of 10 kΩ accidentally grasps the terminals of a 14 kV power supply. (a) If the internal resistance of the power supply is 2000 Ω, what is the current through the person’s body? (b) What is the power dissipated in his body? (c) If the power supply is to be made safe by increasing its internal resistance, what should the internal resistance be for the maximum current in the above situation to be 1.00 mA or less?

Rperson Rint

V

Power supply

September 23, 2011

26-8 Three resistors having resistances of R1 = 1.60 Ω, R2 = 2.90 Ω, and R3 = 4.60 Ω are connected in parallel to a 26.0 V battery that has negligible internal resistance. (a) Find the equivalent resistance of the combination. (b,c,d) Find the current in through each resistor. (e) Find the total current through the battery. (f,g,h) Find the voltage across each resistor. (i,j,k) Find the power dissipated through each resistor. (l) Which resistor dissipates the most power: the one with the greatest resistance or the least resistance? (a) The equivalent resistance for resistors in parallel is given by the sum of their reciprocals, 1 1 1 1 = + + . Req R1 R2 R3

(a) Write the loop and node equations needed to determine the currents I1 , I2 , and I3 in the circuit shown. Indicate clearly the loop used to determine each loop equation. node: loop 1: loop 2:

I2 = I1 + I3 9 − I1 − 2I1 − 4I2 = 0 8 − 2I3 − 1 − 4I2 = 0

(b) Determine the currents by explicit solution of the equations. You must show your work. Rearrange loop 1 and loop 2 equations: loop 1: loop 2:

3I1 + 4I2 = 9 4I2 + 2I3 = 7

Use the node equation to eliminate I3 = I2 − I1 :

Thus, Req = 0.842 Ω. (b,c,d) The voltage is the same across parallel resistors, therefore V = IR ⇒ I = V /R. Thus, V I1 = = 16.3 A; R1

V I2 = = 8.97 A; R2

V I3 = = 5.65 A. R3

(e) The current from the battery is the sum of the currents through each resistor: Itotal = I1 + I2 + I3 = 30.9 A One could also use the equivalent resistance, Itotal = V /Req = 26 V/(0.842 Ω) = 30.9 A. (f,g,h) The voltage is the same across resistors in parallel. V1 = V2 = V3 = 26.0 V (i,j,k) Any formulation of P = V I = I 2 R = V 2 /R will work, P1 = 423 W,

P2 = 233 W,

P3 = 147 W.

(l) P = V 2 /R. Since the voltage is the same across all the resistors, the power is greatest in the smallest resistor. OE 41-1 For this problem, you must write and then solve the loop and node equations needed to find the currents I1 , I2 , and I3 shown in the figure. (a) In the circuit shown in the figure, what is the value of the current I1 ? Remember that I1 may be positive or negative. (b) What is the value of the current I2 ? Remember that I2 may be positive or negative. (c) What is the value of the current I3 ? Remember that I3 may be positive or negative. I1

I3

9V

8V I2

1Ω

4Ω

2Ω

1

2Ω

2

1V

3I1 + 4I2 = 9 −2I1 + 6I2 = 7

(1) (2)

Multiply (1) by 2 and (2) by 3; add and solve for I2 . Substitute back for I1 then I3 . Results are I1 = 1.0 A,

I2 = 1.5 A,

I3 = 0.5 A

Physics 21 Fall, 2011

Solution to HW-9

Set up the loop equation for this new loop: 25 V − (50 Ω)I = 0 =⇒ I =

26-27 In the circuit shown in the figure the batteries have negligible internal resistance and the meters are both idealized. With the switch S open, the voltmeter reads 15 V. (a) Find the emf E of the battery. (b) What will the ammeter read when the switch is closed?

(a) For this part we can ignore the battery with the switch, since the switch is open and there will be no current through it. So let’s draw the circuit that we are concerned with, showing also the currents in each branch and the loops we will use to write Kirchhoff’s equations:

25 V = 0.50 A 50 Ω

26-28 In the circuit shown in the figure both batteries have insignificant internal resistance and the idealized ammeter reads 1.60 A in the direction shown. (a) Find the emf E of the battery. (b) Is the polarity shown correct?

(a) To solve this problem, we will use Kirchhoff’s rules, but this time instead of solving for three currents, we will solve for two currents and the emf. We assign currents I1 and I2 to the remaining branches and loops as shown in the diagram:

We can find I3 since we were given the measured voltage across the 50 Ω resistor. Using Ohm’s Law, V = IR =⇒ I3 =

V 15 V = = 0.30 A. R 50 Ω

Now, let’s write out the loop and node equations: Node : Left loop :

I1 = I2 + I3 E − (20 Ω)I1 − (75 Ω)I2 = 0

Right loop : E − (20 Ω)I1 − (30 Ω)I3 − (50 Ω)I3 = 0 Since we already know I3 , there are three unknowns in this system, E, I1 , and I2 . Here are the equations after substituting for I3 and combining some terms: Node :

I1 = I2 + 0.3

The left loop equation is 0 = 48.0 Ω(I2 ) + 12 Ω(1.6 A)− 75.0 V = 48.0 Ω(I2 ) − 55.8 V, and the right loop equation is 0 = −15.0 Ω(I1 ) + E − 48.0 Ω(I2 ) We can solve the left loop equation directly for I2 : 55.8 V = 1.1625 A. 48.0 Ω

Now substitute this I2 into the node equation to solve for I1 :

Now we have three equations for three unknowns. Next we simplify the equations and solve for E. The solution is I1 = 0.62 A,

I2 = I1 + 1.60 A.

I2 =

Left loop : E − 20I1 − 75I2 = 0 Right loop : E − 20I1 − 24 = 0

E = 36.4 V,

The node equation is

I2 = 0.32 A.

(b) The circuit is different for this part and is shown below. Conveniently, to find the new current in the ammeter, we only need to consider the one loop shown.

I1 = I2 − 1.60 A = −.4375 A Substituting the currents found above into the right loop equations gives 0 = − 15.0 Ω(−.4375 A) + E − 48.0 Ω(1.1625 A) E = 15.0 Ω(−.4375 A) + 48.0 Ω(1.1625 A) = 49.24 V. (b) The polarity of the battery as shown is correct, because the E we calculated was positive. If the calculated E had been negative, it would imply that the assumed polarity in the drawing was incorrect.

September 30, 2011

26-40 A 12.8µF capacitor is connected through a 0.890 MΩ resistor to a constant potential difference of 60.0 V. (a) Compute the charge on the capacitor at the following times after the connections are made: 0 s, 5.0 s, 10.0 s, 20.0 s, and 100.0 s. (b) Compute the charging currents at the same instants. (a) As derived, the formula for the charge on a charging capacitor as a function of time is: ´ ³ q(t) = Qf 1 − e−t/RC ,

26-48 In the circuit shown below, C = 5.90 µF, E = 28.0 V, and the emf has negligible resistance. Initially the capacitor is uncharged and the switch S is in position 1. The switch is then moved to position 2, so the capacitor begins to charge. (a) What will be the charge on the capacitor a long time after the switch is moved to position 2? (b) After the switch has been moved to position 2 for 3.00 ms, the charge on the capacitor is measured to be 110 µC. What is the value of the resistance R? (c) How long after the switch is moved to position 2 will the charge on the capacitor be equal to 99.0% of the final value found in part (a)?

where the final charge Qf = CV = 7.68 × 10−4 C. The time constant τ = RC = 11.392 s. The table below gives q(t) at the times specified. (b) The relationship between charge and current is i = dq/dt, so we can determine the current as a function of time by differentiating the expression for q(t) above: i(t) =

dq Qf −t/RC = e = I0 e−t/RC dt RC

where we substituted Qf = CV and noted that the initial current I0 that flows is the battery voltage V divided by the resistance R. I0 is 6.74 × 10−5 A. Here is a table of the charge and current at various times. Note that 100 s is about nine times the time constant, and at that point the capacitor is essentially fully charged, and the current from the battery is essentially zero. t (s) 0 5 10 20 100

q(t) (C) 0 2.73 × 10−4 4.49 × 10−4 6.35 × 10−4 7.68 × 10−4

i(t) (A) 6.74 × 10−5 4.35 × 10−5 2.80 × 10−5 1.16 × 10−5 1.00 × 10−8

26-43 An emf source with a magnitude of E = 120 V, a resistor with a resistance of R = 87.0 Ω, and a capacitor with a capacitance of C = 3.90 µF are connected in series. As the capacitor charges, when the current in the resistor is 0.700 A, what is the magnitude of the charge on each plate of the capacitor? (a) The simplest way is to apply the loop equation. Let q(t) and i(t) be the instantaneous charge on the capacitor and current in the circuit. Then E − VR − VC = 0 q(t) E − i(t)R − = 0. C Now we can solve for the charge on the capacitor as a function of the current: h i q(t) = C[E − i(t)R] = 3.90 µC 120 V − (0.7 A)(87 Ω) = 230 µC.

(a) We know that after a long time the circuit will approach a steady state where the charge on the capacitor will be simply given by Q = CE. Substituting C = 5.90 µF and E = 28.0 V we find the charge on the capacitor after a long time will be Q = 165.2 µC. (b) We know that the resistance R is part of the time constant in the function q(t). We found q(t) by solving the differential equation obtained from Kirchoff’s loop equation. t

q(t) = CE(1 − e− RC ) Since we know q(t = 3 s) we can rearrange this equation to solve for R. 1 t R=− q C ln(1 − CE ) We insert our value for q(t = 3 s) = 110 µC and find the resistance R = 464 Ω. (c) We want to find at what time t will the charge on the capacitor be 99% of its final value which we found in part (a). In other words we want to solve for t when q = 0.99 × CE. We rearrange our equation for q(t) to get: t = −RC ln(1 −

q ). CE

Substituting our values of R, C, and q we find: t = −(464 Ω)(5.9 × 10−6 F) ln(1 − 0.99) = 0.0126 s

Physics 21 Fall, 2011

Solution to HW-10

27-3 In a 1.35 T magnetic field directed vertically upward, a particle having a charge of magnitude 8.90 µC and initially moving northward at 4.72 km/s is deflected toward the east. (a) What is the sign of the charge of this particle? (b) Find the magnetic force on the particle.

The magnetic force on a charged particle moving in a magnetic field is given by the equation F = qv × B Since, in this case, F, v, and B are mutually perpendicular, the magnitude of F is simply given by F = qvB, with the direction determined by the right hand rule. (a) We need to apply the right hand rule to see if the direction of the force is consistent with a positive charge or a negative charge. Imagine you are seated so that north is in front of you. The other directions are then determined: east is to the right, south is behind you, and west is to the left. So northward velocity means the particle is moving forward. Point the fingers of your right hand straight forward. The magnetic field is upward, so curl the fingers of your right hand upward. In order to do this, your palm must be facing upward. Then, the thumb of your right hand is pointing to the right (eastward). Eastward is the direction the particle is deflected. Thus, the particle must have a positive charge. Here is a diagram: N v

B

F

W

E

27-4 A particle with mass m = 1.81×10−3 kg and a charge of q = 1.22 × 10−8 C has, at a given instant, a velocity v = (3.00 × 104 m/s) ˆj. (a) What is the magnitude of the particle’s acceleration produced by a uniform magnetic field B = (1.63 T) î + (0.980 T) ˆj? (b) What is the direction of the particle’s acceleration? To determine the acceleration of the particle we need to know what force is acting on it. We can assume the only force is due to the magnetic field. The force on a charged particle moving in a magnetic field is given by: F = q (v × B) For the cross product, we notice that many components of v and B are zero: v = vyˆj and B = Bxî + Byˆj. We can evaluate the cross product v × B by using the cross product of each pair of unit vectors: ˆj × î = −k ˆ

ˆj × ˆj = 0

Then í ³ h ˆ F = qv × B = q vyˆj × Bxî + Byˆj = −qvy Bx k Substituting the given quantities we get: F = −(1.22 × 10−8 C)(3.00 × 104 m/s)(1.63 T) ˆ = −5.97 × 10−4 N k Using Newton’s second law, we can find the acceleration of the particle: ˆ a = F/m = −0.330 m/s k.

S

(b) F = qvB = (8.90 µC)(4.72 km/s)(1.35 T) = 0.0567 N

September 30, 2011

27-21 A deuteron (the nucleus of an isotope of hydrogen) has a mass of 3.34 × 10−27 kg and a charge of 1.60 × 10−19 C. The deuteron travels in a circular path with a radius of 6.90 mm in a magnetic field with a magnitude of 2.60 T. (a) Find the speed of the deuteron. (b) Find the time required for it to make 21 of a revolution. (c) Through what potential difference would the deuteron have to be accelerated to acquire this speed?

Note that Mastering Physics is only asking us to provide |V |, so we don’t have to worry about getting the signs right. K0 + U0 = K1 + U1 K1 − K0 = U0 − U1 2 1 2 mv⊥ = q (V0 − V1 ) = −q∆V ¡ ¢¡ ¢ 2 3.34 × 10−27 kg 8.594 × 105 m/s mv⊥ ∆V = − =− 2q 2 (1.60 × 10−19 C) |V | = 7709 V = 7.71 kV 27-25 An electron in the beam of a TV picture tube is accelerated by a potential difference of 1.95 kV. Then it passes through a region of transverse magnetic field, where it moves in a circular arc with a radius of 0.179 m. What is the magnitude of the field? From problem 27-24, we know that B = mv/(eR), but we have to find v for an electron accelerated through a potential difference of ∆V = 1950 volts. By definition, the electron gains an energy 1950 eV. Assuming the electron starts from rest, we have 2 1 2 mv

(a) As is seen in the figure, the force on a moving charged particle due to a magnetic field causes it to travel in a circular path. We have an equation from the equation sheet that describes this motion. R=

mv⊥ qB

Here, the speed is written as v⊥ to remind us that only the component of the velocity vector that is perpendicular to the magnetic field direction contributes to the circular motion. In this problem, we are told it is circular motion, so we know that the velocity is purely perpendicular to the magnetic field direction. We are given numbers for all of the other quantities in this equation, so simply solve it for v⊥ . ¡ ¢¡ ¢ 6.90 × 10−3 m 1.60 × 10−19 C (2.60 T) RqB = v⊥ = m 3.34 × 10−27 kg = 8.594 × 105 m/s (b) For this part, we must recall concepts from Physics 11. We are looking for the time t to complete half a revolution, or to travel a distance d of half the circumference of the circular orbit. Since d = πR, and the speed v⊥ of the particle in its circular orbit is constant, we have πR = v⊥ t =⇒ t = πR/v⊥ Then ¡

−3

¢ π 6.90 × 10 m πR = = 2.52 × 10−8 s = 25.2 ns t= v⊥ 8.594 × 105 m/s (c) This type of problem can be solved by conservation of energy. See example 23.7 in the textbook for a reminder. The particle’s gain in kinetic energy is equal to its loss in potential energy. In this case, the particle starts from rest.

= e∆V = (1950 eV) × 1.602 × 10−19 J/eV

Solving for v gives v = 2.62 × 107 m/s. Then B=

mv = 8.32 × 10−4 m. eR

OH11-05 An electron enters a uniform magnetic field with magnitude 0.3 T at a 45◦ angle to B. Determine the radius r and pitch p (distance between loops) of the electron’s helical path, assuming its speed is 2 × 106 m/s.

Since v makes an angle of 45◦ with B, the components of v parallel and perpendicular to B (vk and v⊥ ) both have the √ √ value v0 / 2 (since sin θ = cos θ = 1/ 2 for θ = 45◦ ), where v0 is the given speed 2.0 × 106 m/s. We get the radius of the circular motion using v⊥ : √ (9.11 × 10−31 )(2.0 × 106 / 2) me v ⊥ = = 26.8 µm r= eB (1.6 × 10−19 )(0.30) The time to make one loop of circular motion is ∆t =

2πr 2πme v⊥ 2πme = = . v⊥ eBv⊥ eB

The pitch is the distance travelled parallel to B (at speed vk ) in time ∆t (one loop): p = vk ∆t =

2πme vk = 1.69 × 10−4 m = 169 µm eB

Physics 21 Fall, 2011

Solution to HW-12

27-36 A straight, vertical wire carries a current of I = 1.19 A downward in a region between the poles of a large superconducting electromagnet, where the magnetic field B has a magnitude of B = 0.591 T and is horizontal. What are the magnitude and direction of the magnetic force on a section of the wire with a length of l = 1.00 cm that is in this uniform magnetic field, if the magnetic field direction is (a) east, (b) south, (c) 34◦ south of west?

27-39 A thin, 51.0 cm long metal bar with mass 770 g rests on, but is not attached to, two metallic supports in a uniform magnetic field with a magnitude of 0.470 T, as shown in the figure. A battery and a resistor of resistance 26.0 Ω are connected in series to the supports. (a) What is the largest voltage the battery can have without breaking the circuit at the supports? (b) The battery voltage has the maximum value calculated in part a. If the resistor suddenly gets partially short-circuited, decreasing its resistance to 2.70 Ω, find the initial acceleration of the bar.

The diagram shows the view looking down in the direction of the current, for an arbitrary orientation of B in the horizontal plane:

N B W

I

S

E Fmag

We use the right hand rule to evaluate the magnetic force Fmag = Il × B, where l is a vector of length l, the length of the wire section, that points in the direction of the current. One can see that for any horizontal B, the magnetic force will also be horizontal, and its direction will be 90◦ clockwise from B. (a) For B pointing east, the magnitude of Fmag is Fmag = IlB = (1.19 A)(0.010 m)(0.591 T) = 7.03 mN and the direction is south. (b) For B pointing south, Fmag is the same as in part (a), and the direction of Fmag is west. (c) For B pointing east, Fmag is the same as in part (a), and the direction of Fmag is 34◦ west of north.

The voltage V causes a current I = V /R to flow in the wire. Then the magnetic field exerts a force on the currentcarrying wire that is in the upward direction. (a) The largest force that won’t break the circuit occurs when the upward magnetic force just balances the downward gravitational force: V Fmag = IL × B = ILB = LB = mg, R where I is the current through the wire, and L is a vector that points along the portion of the wire that is in the field. Solving for V , we find V =

mgR = 819 V. LB

(b) When the resistance R drops to a lower value R , the current in the wire increases, and the upward force on the wire will exceed the downward gravitational force, leading to a net upward force ma: V LB − mg = ma. R Substituting the expression for V derived in part (a) into the expression above leads to mgR LB R − mg = ma ⇒ g − g = a LB R R R − R 2 a=g = 84.6 m/s . R

October 10, 2011

27-42 The plane of a rectangular loop of wire with a width of 5.0 cm and a height of 8.0 cm is parallel to a magnetic field of magnitude 0.16 T. The loop carries a current of 6.3 A. (a) What torque acts on the loop? (b) What is the magnetic moment of the loop? (c) What is the maximum torque that can be obtained with the same total length of wire carrying the same current in this magnetic field?

27-74 A wire 28.0 cm long lies along the z axis and carries a current of 8.60 A in the +z direction. The magnetic field is uniform and has components Bx = −0.200 T, By = −0.968 T, and Bz = −0.327 T. Find the x, y, and z components of the magnetic force on the wire, and the magnitude of that force. The force on the wire is given by

(b) The magnetic moment of the loop µ has magnitude IA and direction perpendicular to the loop. The magnitude is µ = 2.52 × 10−2 A m2 (a) The torque on the loop is τ = µ × B. Because µ is perpendicular to B, the magnitude of τ is τ = µB = 4.03 × 10−3 N m (c) The maximum torque would be achieved for the loop with the largest area for the available length of wire. The shape of the optimum loop is a circle. If the sides are a and b, then the radius r of the circle satisfies 2πr = 2(a + b)

⇒

r=

a+b = 0.13 m. π

The magnitude of the maximum torque is τmax = πr2 IB = 5.42 × 10−3 N m. 27-43 In the Bohr model of the hydrogen atom, in the lowest energy state the electron orbits the proton at a speed of v = 2.2 × 106 m/s in a circular orbit of radius r = 5.3 × 10−11 m. (a) What is the orbital period of the electron? (b) If the orbiting electron is considered to be a current loop, what is the current I? (c) What is the magnetic moment of the atom due to the motion of the electron? (a) The time T to travel distance 2πr at speed v is T = 2πr/v =

2π(5.3 × 10−11 m) = 1.5 × 10−16 s 2.2 × 106 m/s

(b) Current is the amount of charge ∆Q passing a point in time ∆t. Since the electron with charge e passes any point on its orbit once in time T , I = e/T =

1.602 × 10−19 C = 1.1 mA 1.5 × 10−16 s

(c) The magnitude of the magnetic moment of a current loop is the current I times the area of the loop: µ = πr2 I = π(5.3 × 10−11 m)2 (.00106 A) = 9.3 × 10−24 A m2

ˆ × (Bxî + Byˆj + Bz k). ˆ F = IL × B = ILk ˆ × î = ˆj, We can work out the cross products using k ˆ ˆ ˆ ˆ ˆ k × j = −i, and k × k = 0. We get F = IL −Byî + Bxˆj ⇒ Fx = 2.33 T, Fy = −0.482 T F = Fx2 + Fy2 + 02 = 2.38 T

Physics 21 Fall, 2011

Solution to HW-13

28-12 Two parallel wires are 5.00 cm apart and carry currents in opposite directions, as shown in the figure. Find the magnitude and direction of the magnetic field at point P due to two 1.50-mm segments of wire that are opposite each other and each 8.00 cm from P .

28-18 Two long, straight wires, one above the other, are separated by a distance 2a and are parallel to the x axis. Let the +y axis be in the plane of the wires in the direction from the lower wire to the upper wire. Each wire carries current I in the +x direction. Find B, the net magnetic field of the two wires at the following points in the plane of the wires: (a) midway between the wires, (b) at a distance a above the upper wire, and (c) at a distance a below the lower wire.

(b) I

dl

y

2a

(a) I

r-r' dl

r-r'

z

x

(c)

The magnitude of the magnetic field near a long wire is B=

Use the Biot-Savart Law: dB =

µ0 Idl × (r − r′ ) , 3 4π |r − r′ |

where dl points in the direction of the current I, and r − r′ points from the current point to the field point (P ). We can evaluate dBtop and dBbottom (the contributions to the total field from the top and bottom wires, respectively) separately and then add them. The diagram shows the vectors dl and r−r′ for each of these contributions. The simplest way to evaluate the cross product is to use the right hand rule for the direction, and to get the magnitude from the products of the magnitudes |dl| and |r − r′ | and the sin of the angle between the vectors. By the right hand rule, both dBtop and dBbottom point into the page. The magnitudes of dl and r − r′ are 0.0015 m and 0.08 m, respectively, in each case. Since the wires are 5.0 cm apart, the sines of the angles in each case is 2.5/8.0. Hence (12.0)(0.0015)(0.080)(2.5/8.0) 10−7 0.083

dBtop

=

dBbottom

= 8.79 × 10−8 T (24.0)(0.0015)(0.080)(2.5/8.0) = 10−7 0.083 −7 = 1.76 × 10 T.

Since both contributions to the field point into the page, so does the sum, and dBtotoal = dBtop + dBbottom = 2.64 × 10−7 T.

µ0 I , 2πR

where R is the perpendicular distance to the wire. The direction of the field is given by the right hand rule. For the present problem, we must add the vector contributions to the field from each wire at each point (a), (b), and (c). The distances will be multiples of a. (a) At point (a), midway between the wires, the field from the top wire is µ0 I/(2πa) into the page, and the field from the bottom wire is µ0 I/(2πa) out of the page. The vector sum is zero. B=0 (b) At point (b), a distance a above the top wire, the field from the top wire is µ0 I/(2πa) out of the page, and the field from the bottom wire is µ0 I/(2π3a) out of the page. The sum is µ ¶ 2µ0 I 1 µ0 I 1 = + (out of the page). B= 2π a 3a 3πa In terms of vector components, B=

2µ0 I ˆ k. 3πa

(c) At point (c), a distance a below the bottom wire, the field from the bottom wire is µ0 I/(2πa) into the page, and the field from the top wire is µ0 I/(2π3a) into the page. B=

2µ0 I 3πa

(into the page).

In terms of vector components, B=−

2µ0 I ˆ k. 3πa

October 12, 2011

Physics 21 Fall, 2011

Solution to HW-14

Cancelling Magnetic Field Four very long, currentcarrying wires in the same plane intersect to form a square with side lengths of 39.0 cm, as shown in the figure. The currents running through the wires are 8.0 A, 20.0 A, 10.0 A, and I. Find the magnitude and direction of the current I that will make the magnetic field at the center of the square equal to zero.

vector sum of the fields can only be zero on the line between the wires. The magnitude of the field B a distance R from a long wire with current I is B = µ0 I/(2πR). The vector field will be zero at a point on the line between the wires a distance x from the left wire and L − x from the right wire, where the magnitudes of the fields are equal. Then µ0 I2 I1 I2 µ0 I1 = =⇒ = 2πx 2π(L − x) x L−x Solving this equation and substituting values leads to µ ¶ ¶ µ I1 23 x= 38 cm = 8.7 cm L= I1 + I2 23 + 78 MasteringPhysics asks for the distance from the 78 A wire in the direction of the 23 A wire, which is L − x = 29.3 cm in our notation. (b) Panel (b) of the diagram is similar to panel (a); it shows the two wires and the fields when the currents are in opposite directions. In this case the vector sum of the fields can only be zero on the line connecting the wires, but outside the wires. We consider first a distance x to the left of I1 ; in this case the condition is µ0 I2 I1 I2 µ0 I1 = =⇒ = 2πx 2π(L + x) x L+x

The field point at the center of the square is equidistant from all four wires. Let this distance be d = 21 × 0.39 m. We just have to keep track of the direction of each field using the right hand rule. Let out of the page be plus, and let I > 0 correspond to up: B out of page =

µ0 (−10 + I − 8 + 20) 2πd

Solving, we get I = −2 A, and the minus sign means I is directed downward. 28-22 Two long, parallel transmission lines, L = 38.0 cm apart, carry currents I1 = 23.0-A and I2 = 78.0-A. Find all locations where the net magnetic field of the two wires is zero if these currents are (a) in the same direction or (b) in opposite directions. (b)

(a) L x I1

L

x I2

I1

I2

(a) Panel (a) of the diagram shows the two wires end on when the currents are in the same direction. L is the distance between the wires. The magnetic field lines due to each wire separately are shown by the concentric circles (dashed for I1 , solid for I2 ). The direction of the field follows from the right hand rule and is shown at selected points by an arrow next to each circle. By looking at the directions of the two fields in various locations, it’s easy to see that for case (a), the

Solving this equation and substituting values leads to µ ¶ ¶ µ I1 23 x= 38 cm = 15.9 cm L= I2 − I1 78 − 23 MasteringPhysics asks for the distance from the 78 A wire in the direction of the 23 A wire, which is L + x = 53.9 cm in our notation. What about a point to the right of I2 in panel (b)? We can set up the equation; a point a distance x to the right of I2 would be x + L from I1 . Then ¶ µ I1 I2 I2 L = =⇒ x = x L+x I1 − I2 This solution won’t work in our case. I2 > I1 , so x < 0, which contradicts our initial assumption that x is a positive distance. Therefore there are no other solutions. Comparing the two solutions we have obtained for part (b), one can see that a general way of writing the solution is ¶ µ I< L, x= I> − I< where I< (I> ) is the lesser (greater) of I1 and I2 . The point where the field is zero is outside the two currents, a distance x from the wire with the smaller current.

October 14, 2011

Wire and Square Loop A square loop of wire with side length a carries a current I1 . The center of the loop is located a distance d from an infinite wire carrying a current I2 . The infinite wire and loop are in the same plane; two sides of the square loop are parallel to the wire and two are perpendicular as shown. (a) What is the magnitude F of the net force on the loop? (b) The magnetic moment µ of a current loop is defined as the vector whose magnitude equals the area of the loop times the magnitude of the current flowing in it (µ = IA), and whose direction is perpendicular to the plane in which the current flows. Find the magnitude F of the force on the loop from Part (a) in terms of the magnitude of its magnetic moment.

(a) The B field is into the page everywhere on the right of the wire in the plane of the square loop. Its magnitude is given by B = µ0 I2 /2πR, where R is the distance to the wire. The diagram shows the forces on the left and right hand side of the loop from this field. These forces are obtained from F = Il × B, and the magnitudes are µ0 I2 2π(d − a/2) µ0 I2 = I1 aB = I1 a 2π(d + a/2)

Fleft = I1 aB = I1 a Fright

There is also an upward force (Ftop ) and (Ftop ), but they are equal in magnitude and oppositely directed. Hence the net force is the vector sum of the forces to the left and to the right. The net force is to the left and has magnitude µ ¶ 1 1 µ0 I1 I2 a − Fleft − Fright = 2π d − a/2 d + a/2 µ0 I1 I2 a2 = 2 2π d − a2 /4 (b) The magnitude of the magnetic moment µ of the loop is the current times the area, or 2

µ = I1 a . We can write the net force in terms of µ as Fleft − Fright =

µI2 µ0 2π d2 − a2 /4

28-26 Two long, parallel wires are separated by a distance of d = 2.70 cm. The force per unit length that each wire exerts on the other is 4.10 × 10−5 N/m, and the wires repel each other. The current in one wire is I1 = 0.700 A. (a) What is the current in the second wire? (b) Are the two currents in the same direction or in opposite directions? I1

B1 I2

Let’s assume that the currents flow in the directions shown, and we’ll show that the force between the wires is repulsive. The magnitude of the magnetic field B1 of wire 1 at wire 2 is µ0 I1 , B1 = 2πd and from the rh rule, B1 points into the page. The magnitude of the force F that B1 exerts on a length L of wire 2 is µ0 I1 F = I2 LB1 = I2 L , 2πd and by the rh rule one can see that the direction is downward. One can go through a similar argument to find that the force on the upper wire has the same magnitude and is upward, so the forces make the wires repel each other. The force per unit length is F µ0 I1 I2 = L 2π d The question asks for I2 , so µ ¶ ¡ ¢ 0.027 2π F d = 0.5×107 4.10× 10−5 = 7.91 A I2 = µ0 L I1 0.7 In general, the currents must flow in the opposite direction. 28-27 The wires in a household lamp cord are typically d = 2.5 mm apart center to center and carry equal currents in opposite directions. (a) If the cord carries current to a 100 watt light bulb connected across a 120 V potential difference, what force per meter does each wire of the cord exert on the other? (Model the lamp cord as a very long straight wire.) (b) Is the force attractive or repulsive? (c) Is this force large enough so it should be considered in the design of lamp cord? Since P = IV , the current in each wire is I = P/V = 100 W/120 V = 0.833 A. (a) The force per unit length is given by the formula derived in problem 28-26, with both currents equal: F µ0 I 2 (0.833)2 = = 2 × 10−7 = 5.56 × 10−5 N/m L 2π d 0.0025 (b) The force will be repulsive. (c) No. The force is small compared to the gravitational force. If we guess that a meter of wire weighs a few ounces, say 0.1 kg, then 2

mg = (0.1 kg)9.81 m/s ∼ 1 N, much larger than the magnetic force.

Physics 21 Fall, 2011

Solution to HW-15

28-32 A solid conductor with radius a is supported by insulating disks on the axis of a conducting tube with inner radius b and outer radius c. The central conductor and tube carry equal currents I in opposite directions. The currents are distributed uniformly over the cross sections of each conductor. (a) Derive an expression for the magnitude of the magnetic field at points outside the central, solid conductor, but inside the tube (a < r < b). (b) Derive an expression for the magnitude of the magnetic field at points outside the tube (r > c).

28-34 A closely wound coil has a radius a = 5.90 cm and carries a current I = 3.30 A. How many turns must it have if, at a point on the coil axis 6.20 cm from the center of the coil, the magnetic field is 6.58 × 10−4 T? The magnitude of the magnetic field at the point x on the axis of a single circular loop with a radius a and current I is B=

µ0 Ia2 . 2(x2 + a2 )3/2

For a loop with N turns, the total magnetic field will be N times this magnitude. So we can determine the number of turns by solving for N to get: 2B(x2 + a2 )3/2 µ0 Ia2 ¢3/2 ¡ 2(.000658) .0622 + .0592 = 57 = (4π × 10−7 )(3.30)(.059)2

N=

(a) This problem is best answered with Ampere’s Law I B · dl = µ0 Iencl Chose a circular line integral path with radius r between the center and outer conductor, then by symmetry we expect the value of B to be constant around the path. The only current enclosed is the current I in the center conductor. 2πrBinside = µ0 I Solving for B, Binside =

¡ ¢I µ0 I = 2 × 10−7 2πr r

(b) Taking a similar approach for this problem, choose a current loop with radius r > c. The total current enclosed is now zero because the two currents I in each conductor are going in opposite directions. Boutside = 0

October 18, 2011

28-36 The figure shows, in cross section, several conductors that carry currents through the plane of the figure. The currents have the magnitudes I1 = 4.0 A, I2 = 6.5 A, and I3 = 2.1 A, and the directions shown. Four paths,H labeled a, b, c, and d, are shown. What is the line integral B · dl for each of the four paths? The integral involves going around the path in the counterclockwise direction.

28-41 A solenoid is designed to produce a magnetic field of 2.00 × 10−2 T at its center. It has a radius of 1.60 cm and a length of 33.0 cm, and the wire can carry a maximum current of 13.5 A. (a) What minimum number of turns per unit length must the solenoid have? (b) What total length of wire is required? (a) The magnitude of the magnetic field inside a solenoid is given by B = µ0 nI, where n is the number of turns per unit length. Solving this equation for n = B/µ0 I, we can see that the minimum n will occur when the current is at its maximum; n=

turns 2.00 × 10−2 T B = 1180 = . Wb −7 µ0 Imax m (4π × 10 A·m )(13.5 A)

(b) The total length of wire L will be the total number of loops times the length of one loop, which is the circumference, L = N 2πr. The total number of loops is simply the number of turns per length times the total length of the solenoid N = nl. Thus the total length of the wire is L = nl2πr = (1180 m−1 )(.33 m)(2π)(.016 m) = 39.1 m

Use Ampere’s Law, I

B · dl = µ0 Iencl ,

where the positive direction of current flow is out of the page since the integral is done counterclockwise. (a) I Iencl = 0 ⇒

28-45 A wooden ring whose mean diameter is 16.0 cm is wound with a closely spaced toroidal winding of 585 turns. (a) Compute the magnitude of the magnetic field at the center of the cross section of the windings when the current in the windings is 0.655 A.

B · dl = 0

(b) Iencl = −I1 ⇒

I

B · dl = −5.03 × 10−6 T m.

(c) Iencl = I2 − I1 ⇒

I

B · dl = 3.14 × 10−6 T m.

(d) Iencl = I2 + I3 − I1 ⇒

I

B · dl = 5.78 × 10−6 T m. (a) In the center of the windings, the strength of the magnetic field can be found using the formula derived in the class lecture: B = µ0 nI This is the same formula used for a solenoid; a toroid is basically a rolled up solenoid. n is the number of turns per unit length along the toroid. To calculate n, we use the number of turns divided by the circumference of the toroid: n= B = µ0

585 π 0.16

585 0.655 = 9.58 × 10−4 T π 0.16

Physics 21 Fall, 2011

Solution to HW-16

F =

28-48 The current in the windings of a toroidal solenoid is 2.400 A. There are N = 500 turns and the mean radius is r = 25.00 cm. The toroidal solenoid is filled with a magnetic material. The magnetic field inside the windings is found to be 1.940 T. (a) Calculate the relative permeability. (b) Calculate the magnetic susceptibility of the material that fills the toroid. (a) The magnetic field inside a tightly wound toroidal solenoid is µ0 N I B = Km µ0 nI = Km , 2πr where n is the number of turns per unit length and N is the total number of turns. Solving the last equation for Km , we get Km =

Using SI units,

2πrB 2π · 0.25 · 1.94 = = 2021. µ0 N I (4π × 10−7 ) 500 · 2.4

(b) The magnetic susceptibility is χm = K m − 1 thus the answer is 2020. 28-52 A long, straight wire carries a current of 2.50 A. An electron is traveling in the vicinity of the wire. At the instant when the electron is 4.50 cm from the wire and traveling with a speed of 6.00 × 104 m/s directly toward the wire, what are the magnitude and direction (relative to the direction of the current) of the force that the magnetic field of the current exerts on the electron?

B

F

v

I The magnetic field due to the wire has magnitude B=

µ0 I 2πr

and direction (from the rh rule) out of the page at the location of the electron as shown. The force on the electron is given by F = qv × B = −ev × B.

(1.602 × 10−19 )(60000)(4π × 10−7 )(2.5) N 2π(.045)

= 1.07 × 10−19 N The magnetic field exerts a force in the same direction as the current. 28-55 Two identical circular, wire loops 35.0 cm in diameter each carry a current of 2.30 A in the same direction. These loops are parallel to each other and are 22.0 cm apart. Line ab is normal to the plane of the loops and passes through their centers. A proton is fired at 2750 m/s perpendicular to line ab from a point midway between the centers of the loops. Find the magnitude of the magnetic force these loops exert on the proton just after it is fired. This problem involves the magnetic force on a moving charged particle, F = qv × B. The trick is to find the magnetic field at the position of the particle due to the two loops. The particle is located on line ab. From the description of the loops, we know that line ab is on the axis of both loops. Section 28-5 in the textbook shows how to calculate the magnetic field along the axis of a loop. In particular, for a loop of radius a, carrying a current I, the magnetic field along the axis a distance x from the loop is given by equation 28.15: B=

µ0 Ia2 2 (x2 + a2 )

3/2

The direction of the magnetic field is along the axis as determined by the right hand rule. Notice that the direction of the force is the same regardless of which side of the loop is particle is located at. Since the particle is located equidistant from the two loops, and because the loops carry the same current in the same direction, each loop contributes the same magnitude and direction of magnetic field. The total magnetic field is thus twice the magnetic field due to a single loop. We only need to find the magnitude of the force, F = qvB sin θ = qvB sin 90◦ = qvB, since the direction of v is perpendicular to the loop axis, and the direction of B is along the loop axis. Ã ! µ0 Ia2 µ0 Ia2 F = qv 2 = qv 3/2 3/2 2 (x2 + a2 ) (x2 + a2 ) ¡ ¢ = 1.60 × 10−19 C (2750 m/s) ¡ ¢ 2 4π × 10−7 T · m/A (2.30 A) (.175 m) × ´3/2 ³ 2 2 (.110 m) + (.175 m) = 4.42 × 10−21 N

F is in the direction shown in the diagram. Note that the negative sign of the electron makes F in the opposite direction from v × B. The vectors F, v, and B are mutually perpendicular, so the magnitude of F is F = evB =

evµ0 I 2πr

October 21, 2011

28-63 Two long, parallel wires hang by 4.00-cm-long cords from a common axis (see the figure ). The wires have a mass per unit length of 1.10 × 10−2 kg/m and carry the same current in opposite directions. (a) What is the current in each wire if the cords hang at an angle of 6.00◦ with the vertical?

(a) Two parallel wires with currents running in opposite directions exert a repulsive force on one another. Using the formula for the magnetic field at a distance r from a long wire, B = µ0 I/(2πr), and the force F on a length L of wire from a magnetic field B perpendicular to the wire, F = ILB, one can find the force per unit length on one wire due to the current in the other: µ0 II ′ F = . L 2πr Viewing a wire end-on allows us to construct a free body diagram involving the gravitational force, tension, and the magnetic force.

The sum of the forces can then be written as X Fy = T cos θ − mg = 0 =⇒ T = mg/ cos θ X

Fx = F − T sin θ = 0 =⇒ F = T sin θ = mg tan θ,

where F denotes the force on a length of wire L. Now we use the expression for the magnetic force F and the linear mass density, λ = m/L, µ0 I 2 L = mg tan θ = λLg tan θ 2πr r µ0 I 2 L 1 = λLg tan θ =⇒ I = 2πrλg tan θ 2πr µ0

F =

Note the Ls cancel. The distance r between the two wires is twice the base leg of the right triangle that is formed when the 4.00-cm cord hangs at 6◦ with respect to the vertical. r = 2 l sin θ = 2(0.040 m) sin(6.00◦ ) = 8.362 × 10−3 m. Now we substitute this r and µ0 = 4π × 10−7 T m/A, λ = 1.10 × 10−2 kg/m, g = 9.8 m/s2 , and θ = 6.0◦ into the expression for I and find I = 21.8 A,

When I is increased, the angle θ from the vertical increases. A large current is required for even the small displacement seen here. 29-5 A circular loop of wire with a radius of r = 14.0 cm and oriented in the horizontal xy plane is located in a region of uniform magnetic field. A field of magnitude B = 1.8 T is directed along the positive z direction, which is upward. (a) If the loop is removed from the field region in a time interval of 2.0 ms, find the average of the emf that will be induced in the wire loop during the extraction process. (b) If the coil is viewed looking down on it from above, is the induced current in the loop clockwise or counterclockwise?

(a) The induced emf is given by Faraday’s Law: ¯ ¯ ¯ ∆Φ ¯ dΦ ¯. =⇒ |E| ∼ ¯¯ E =− dt ∆t ¯

The change in flux ∆Φ is the difference between the final flux and the initial flux, and ∆t is 2.0 µs. Since the field B is initially perpendicular to the loop, the initial flux Φi through the loop is just the area of the circular loop times the magnetic field: Z Φi = B · dA = πr2 B. After the loop is removed from the magnetic field, the final flux Φf is zero. We get the absolute value of the emf from ¯ ¯ ¯ ¯ 2 ¯ Φf − Φi ¯ ¯ 0 − Φi ¯ ¯=¯ ¯ = Φi = πr B |E| = ¯¯ ¯ ¯ ¯ ∆t ∆t ∆t ∆t 2 π(0.14 m) × 1.8 T = = 55 V. 0.002 s (b) The direction of the induced current I is obtained using Lenz’s Law. The direction of I must oppose the change in flux. Originally, the flux was determined by the field B pointing up (as one looks down on the loop). When the loop is removed from the field, the induced current will be in the direction that will restore an upward pointing field. Therefore the current is counterclockwise.

Physics 21 Fall, 2011

Solution to HW-17

29-7 The current in the long, straight wire AB shown in the figure is upward and is increasing steadily at a rate di/dt. (a,b) At an instant when the current is i, what are the magnitude and direction of the field B at a distance r to the right of the wire? (c) What is the flux dΦB through the narrow shaded strip? (d) What is the total flux through the loop? (e) What is the induced emf in the loop? (f) Evaluate the numerical value of the induced emf if a = 12.0 cm, b = 36.0 cm, L = 24.0 cm, and di/dt = 9.60 A/s.

29-8 A flat, circular, steel loop of radius 75 cm is at rest in a uniform magnetic field, as shown in an edge-on view in the figure. The field is changing with time, according to −1 B(t) = (1.4 T)e−(0.057s )t . (a) Find the emf induced in the loop as a function of time (assume t is in seconds). (b) When is the induced emf equal to 1/20 of its initial value? (c) Find the direction of the current induced in the loop, as viewed from above the loop.

(a) We can find the emf induced in the loop using I Z d dΦ E · dl = − B · dA = − , dt dt (a,b) B=

µ0 i(t) 2πr

into page (by right hand rule)

(c) µ0 i(t) L dr 2πr

dΦ = B dA = (d) Integrate Φ=

Z

b

a

dΦ =

µ0 i(t)L 2π

Z

a

b

dr µ0 i(t)L b = ln r 2π a

(e) dΦ µ0 L di b |E| = = ln dt 2π dt a (f) (2 × 10

−7

µ

36 )(0.24 m)(9.6) ln 12

¶

= 5.06 × 10−7 V

What will be the direction of the current? Counterclockwise.

where the integral of E · dl is the induced emf. Because the magnetic field makes a 60 degree angle with the loop, we can write Z B · dA = Φ = B(t)A sin(60) where A = πr2 is the area of the loop. Because the area of the loop doesn’t change with time, dΦ/dt = πr2 dB/dt, where d B(t) = (1.4)(−0.057)e−0.057t dt Putting these pieces together results in: E = −πr2 sin(60)(1.4)(−0.057)e−0.057t = 0.122e−0.057t We usually take E to be the magnitude of the emf, and don’t worry about the sign. The negative sign is related to the polarity of the voltage. (b) Now that we have an expression for the induced emf as a function of time, we can solve for t at a particular emf. 1 = e−0.057t =⇒ t = ln 20/(0.057 s−1 ) = 52.6 s 20 (c) According to Lenz’s Law, the induced emf will cause current to flow in the direction that opposes the change in magnetic flux. Because the magnitude of the magnetic field directed upwards through the loop is decreasing with time, the induced emf will cause a current to flow in a direction that will cause a magnetic field upward. For that to happen, the current must flow in a counterclockwise direction.

October 27, 2011

29-11 In a region of space, a magnetic field points in the +x direction (toward the right). Its magnitude varies with position according to the formula Bx = B0 + bx, where B0 and b are positive constants, for x ≥ 0. A flat coil of area A moves with uniform speed v from right to left with the plane of its area always perpendicular to this field. (a) What is the emf induced in this coil while it is to the right of the origin? (b) As viewed from the origin (which we take to be to the left), what is the direction (clockwise or counterclockwise) of the current induced in the coil? (c) If instead the coil moved from left to right, what would be the answer to part (a)? (d) If instead the coil moved from left to right, what would be the answer to part (b)?

29-13 The armature of a small generator consists of a flat, square coil with 120 turns and sides with a length of 1.95 cm. The coil rotates in a magnetic field of 7.80× 102 T. (a) What is the angular speed of the coil if the maximum emf produced is 2.60 × 102 V? (a) The figure below shows the general configuration of the square coil and the magnetic field. Note that our coil has 120 turns. The magnetic field is constant and to the right, while the coil (and consequently, the vector A perpendicular to the loop) changes with time.

B(x) = B 0 + bx +X v

(a) This problem should be solved with Faraday’s Law: E =−

d Φ. dt

The area of the loop remains constant, but the magnetic flux will change with time because the loop moves towards a weaker field. Φ = A (B0 + bx) Using the following equation, we can get the time derivative of the flux at the location x of the loop: dΦ dΦ dx = = (Ab)(−v) = −Abv. dt dx dt Now the emf can be calculated using dΦ = Abv. E =− dt (b) The direction of the current can be determined with Lenz’s Law. The direction of the induced current will be in a direction to oppose the change in flux. Because the flux through the loop is decreasing, the induced current will induce a magnetic field inside the loop pointing to the right. This means, as viewed from the origin (on the left), the current will be going in a clockwise direction (CW). (c) If the coil is moving from left to right instead, the problem is similar to part (a), except the velocity is positive, and so the answer is −Abv. Since no sign convention is defined for the emf enter Abv for Mastering Physics. (d) Because the coil is moving in the opposite direction compared to parts (a) and (b), the flux through the loop is increasing, and so the induced current will induce a magnetic field pointing towards from the origin. So, viewed from the origin (on the left), the current will be in a counter-clockwise direction (CCW), the opposite direction from part (b).

The angle φ between the magnetic field B and the vector A is given by φ = ωt. Thus dφ/dt = ω, and the flux through the coil at any given time can be written as, ΦB = B · A = BA cos φ = BA cos ωt. ¿From Faraday’s law, the induced emf is given by the change of the flux with respect to time. But here, we must remember we have a coil with N = 120 turns, thus, E = −N

dΦB = ωN BA sin ωt dt

The maximum induced emf occurs when sin ωt = 1. We can solve for the angular speed that would create the stated emf: Emax = ωN BA 2.60 × 102 V rad Emax = = 7.31 ω= N BA 120(7.80 × 102 T)(0.0195 m)2 s

29-17 Consider the system shown below. (a) Using Lenz’s law, determine the direction of the current in resistor ab of the figure when switch S is opened after having been closed for several minutes. (b) Using Lenz’s law, determine the direction of the current in resistor ab of the figure when coil B is brought closer to coil A with the switch closed. (c) Using Lenz’s law, determine the direction of the current in resistor ab of the figure when the resistance of R is decreased while the switch remains closed.

29-20 A 1.60 m long metal bar is pulled to the right at a steady 4.5 m/s perpendicular to a uniform, 0.755 T magnetic field. The bar rides on parallel metal rails connected through R = 26.0 Ω, as shown in the figure, so the apparatus makes a complete circuit. You can ignore the resistance of the bar and the rails. (a) Calculate the magnitude of the emf induced in the circuit. (b) Find the direction of the current induced in the circuit. (c) Calculate the current through the resistor.

(a) The current will flow from the positive terminal of the battery when S is closed. Thus, the current flowing in coil A will produce a magnetic field inside coil A pointing to the right (and also to the right inside coil B). When S is then opened, the current will stop flowing and the magnetic field it produces will decrease. Lenz’s Law says that the E induced in the coil B will oppose the change in magnetic flux by causing current that will create an additional magnetic field to the right. Thus, by the right hand rule the current in coil B will flow from a to b. (b) When S is closed, coil A produces a magnetic field inside the coils pointing to the right. As coil B is brought closer to A, the magnetic field inside coil B becomes stronger, so the flux becomes larger. The E induced in coil B will oppose the change in magnetic flux by causing current that will create additional magnetic field to the left. By the right hand rule, the current in coil B will flow from b to a. (c) If the resistance in R is decreased, the relation I = V /R tells us that the current will increase, since the battery voltage V does not change. When the current through coil A increases, the magnetic field directed to the right will also increase. The E induced in coil B will oppose the change in magnetic flux by causing current that will create additional magnetic field to the left. By the right hand rule, the current in coil B will flow from b to a. [This argument is exactly the same as the one used in part (b)].

The flux at any time is the area of the loop times the magnitude of the magnetic field B, and we let L be the length of the bar. In a time ∆t the area of the circuit will increase by L∆x, where ∆x and ∆t are related to the speed of the bar by v = ∆x/∆t. (a) The emf is given by E = ∆Φ/∆t = BL∆x/∆t = BLv = 0.755 T × 1.60 m × 4.5 m/s = 5.44 V (b) Pulling the rod increases the flux by adding magnetic field directed into the page. By Lenz’ Law, the induced current must then produce a magnetic field inside the loop directed out of the page. The current must therefore be counterclockwise. (c) By Ohm’s Law, E = IR ⇒ I = 5.44 V/26.0 Ω = 0.21 A

Physics 21 Fall, 2011

Solution to HW-18

30-7 At the instant when the current in an inductor is increasing at a rate of 6.45 × 10−2 A/s, the magnitude of the self-induced emf is 1.65 × 10−2 V. (a) What is the inductance of the inductor? (b) If the inductor is a solenoid with 405 turns, what is the average magnetic flux through each turn when the current is 0.715 A? (a) In the class lecture we applied Faraday’s Law to a B solenoid with N turns. The self-induced emf is E = −N dΦ dt , and we defined the self-inductance L by E = −L

di dt

Here we aren’t concerned with the direction of the emf, only the magnitude. Therefore we can substitute the given values to obtain 1.65 × 10−2 V E = di/dt 6.45 × 10−2 A/s = 0.256 H

L=

(b) For this part we must notice that since the flux through one loop of a solenoid of length l is given by ΦB = BA = µ0

N ΦB N2 A = N µ0 A = N . l l i

Note that the relation above, L=N

ΦB , i

is in fact the definition used in the book for L [Eq. (30.6)]. We can use it to answer this problem by solving for ΦB . (0.256 H) (0.715 A) Li = N 405 −4 = 4.52 × 10 Wb

ΦB =

30-19 An inductor with an inductance of 2.50 H and a resistance of 8.00 Ω is connected to the terminals of a battery with an emf of 6.00 V and negligible internal resistance. Find (a) the initial rate of increase of current in the circuit; (b) the rate of increase of current at the instant when the current is 0.500 A; (c) the current 0.250 s after the circuit is closed; (d) the final steady-state current. (a) Applying Kirchhoff’s Law, we find the loop equation is E − iR − L

E − iR 6 V − (0.5 A × 8 Ω) di = = = 0.8 A/s. dt L 2.5 H (c) Using the general solution for the loop equation, we have ´ 6V ³ ´ E ³ i= 1 − e−(R/L)t = 1 − e−(8Ω/2.5 H)0.25 s R 8Ω = 0.413 A. (d) The general solution shows that as t → ∞, the exponential term vanishes. Then i=

di E − iR di = 0 =⇒ = . dt dt L

E 6V = = 0.75 A. R 8Ω

30-29 A 7.60 nF capacitor is charged up to 13.0 V, then disconnected from the power supply and connected in series through a coil. The period of oscillation of the circuit is then measured to be 9.00×10−5 s. (a) Calculate the inductance of the coil. (b) Calculate the maximum charge on the capacitor. (c) Calculate the total energy of the circuit. (d) Calculate the maximum current in the circuit. (a) The resonant frequency of an LC circuit is related to L, C, and the period of the oscillations T by

N iA, l

we can rewrite the expression for the inductance L given on the equation sheet in terms of the flux, L = µ0

(b) Using the loop equation again for i = 0.5A,

ω=√

1 2π = . T LC

(The frequency formula is given on the equation sheet; for this case R = 0.) Solving for L leads to L=

(9.00 × 10−5 s)2 T2 = = 2.70 × 10−2 H. 4π 2 C 4π 2 (7.60 × 10−9 F)

(b) The capacitor has its maximum charge when it is initially charged to V = 13 V: Q = CV = (7.60 × 10−9 F)(13 V) = 9.88 × 10−8 C. (c) Energy is conserved, and the total energy of the oscillator is the energy initally stored in the capacitor: U = 21 CV 2 = 21 (7.60 × 10−9 F)(13 V)2 = 6.42 × 10−7 J. (d) In the LC oscillator, the energy swings back and forth between the capacitor and the inductor. At one point, the total energy U of the circuit is all in the inductor. At that time, the current through the inductor is a maximum, and we can use the expression for energy in the inductor, U = 21 LI 2 , and solve for I: r 2U I= L s 2(6.42 × 10−7 J) = 6.90 × 10−3 A = 6.90 mA. = (2.70 × 10−2 H)

The initial condition is that i = 0 at t = 0, so at that time E 6V di = = = 2.4 A/s. dt L 2.5 H

October 28, 2011

30-22 In the figure below, switch S1 is closed while switch S2 is kept open. The inductance is L = 0.115 H, and the resistance is R = 120 Ω. a) When the current has reached its final value, the energy stored in the inductor is 0.260 J. What is the emf E of the battery? b) After the current has reached its final value, S1 is opened and S2 is closed. How much time does it take for the energy stored in the inductor to decrease to half of its original value?

30-33 An LC circuit containing an 86.0 mH inductor and a 1.25 nF capacitor oscillates with a maximum current of 0.760 A. (a) Calculate the maximum charge on the capacitor. (b) Calculate the oscillation frequency of the circuit. (c) Assuming the capacitor had its maximum charge at time t = 0, calculate the energy stored in the inductor after 2.60 ms of oscillation. (a) The charge on the capacitor oscillates as a function of time. Assuming the capacitor starts with the maximum charge at t = 0 (as stated in part c), then we know what the charge and current will look like as functions of time:

(a) When the current has reached its final value I0 , the energy stored in the inductor is given by U0 = 21 LI02 = 12 L

µ

E R

¶2

,

where we used Ohm’s Law to express the current in terms of the emf and resistance. We can rearrange this to solve for the emf and substitute our known values to find E=

µ

2 UR L

¶1 2

2

=

µ

¶1 2(0.260 J)(120 Ω)2 2 = 255 V. (0.115 H)

q(t) = Qmax cos(ωt) dq = −ωQmax sin(ωt) = −Imax sin(ωt) i(t) = dt √ where ω = 1/ LC. The maximum quantities correspond to the charge and current at times when the cosine and sine are equal to one, respectively. Since q(t) and i(t) are related by a time derivative, the second equation above gives a relationship between Qmax and Imax : √ Qmax = Imax /ω = Imax LC p = (0.760 A) (0.086 H)(1.25 × 10−9 F) = 7.88 × 10−6 C (b) The oscillation frequency of the circuit is related to ω by: f=

1 ω 1 p √ = = 2π 2π LC 2π (0.086 H)(1.25 × 10−9 F)

= 1.54 × 10−4 Hz

(b) When the switches are changed, the inductor will release the energy stored in the magnetic field, and this will lead to a current that decays exponentially with time constant L/R (see page 1044 of the text.) The current is given by

(c) The energy stored in the inductor at t′ = 2.60 ms is given by

i(t) = I0 exp(−Rt/L),

= 21 (0.086 H)(0.0760 A)2 £ ¤ × sin2 2π(1.54 × 10−4 hz)(0.0026 s)

where I0 is the steady state current from part (a). We can use this result to determine the energy stored in the inductor as a function of time: U (t) = 21 Li(t)2 = 12 LI02 exp(−2Rt/L) = U0 exp(−2Rt/L). We want to find the time t when the inductor has half of its initial energy. At that time, exp(−2Rt/L) = 21 . Solving this equation for t we find t=

L 0.115 H ln 2 = ln 2 = 0.332 ms. 2R 2 (120 Ω)

U (t′ ) = 21 L[i(t′ )]2 2 = 12 LImax sin2 (2πf t′ )

= 7.03 × 10−3 J

Physics 21 Fall, 2011

Solution to HW-19

29-34 A dielectric of permittivity 3.3 × 10−11 F/m completely fills the volume between two capacitor plates. For t > 0 the electric flux through the dielectric is (7800 V m/s3 )t3 . The dielectric is ideal and nonmagnetic; the conduction current in the dielectric is zero. At what time does the displacement current in the dielectric equal 23 µA?

29-51 As a new electrical engineer for the local power company, you are assigned the project of designing a generator of sinusoidal ac voltage with a maximum voltage of 120 V. Besides plenty of wire, you have two strong magnets that can produce a constant uniform magnetic field of 1.8 T over a square area with a length of 10.4 cm on a side when the magnets are separated by a distance of 12.1 cm. The basic design should consist of a square coil turning in the uniform magnetic field. To have an acceptable coil resistance, the coil can have at most 450 loops.

For a charging capacitor we have the equation for displaceE ment current iD = ǫ0 dΦ dt . When a dielectric is between the capacitor plates, we multiply ǫ0 by the dielectric constant K to get the permittivity of the material ǫ = Kǫ0 . Thus we E have iD = ǫ dΦ dt . The function for the electric flux can then be substituted: ¡ 3¢ ¢ d ¡ 3 3 3 d t (7800 V m/s )t = ǫ(7800 V m/s ) iD = ǫ dt dt = ǫ(7800 V m/s3 )(3)t2 Solving this for time: s iD t= ǫ(7800 V m/s3 )(3) s 23 × 10−6 = 5.5 s = 3.3 × 10−11 (7800 V m/s3 )(3)

We can use Faraday’s law to calculate the induced EMF. I Z d dΦ E · dl = − B · dA = − , dt dt The flux through a square loop perpendicular to the magnetic field with the maximum possible dimensions and N loops of wire will be ΦB = N BA, where A is the area. If the loop is rotated within the magnetic field, then the flux through the loop will change because number of field lines going through the loop will be changing. Rotating at an angular frequency of ω, the flux will change with time: ΦB (t) = N BA cos(ωt) The time derivative of this gives the induced EMF: E =−

dΦB = N BAω sin(ωt) dt

The amplitude is then N BAω. Solving for ω ω=

120 V E = 13.7 rad/sec = N BA 450 × 1.8 × .1042

Mastering physics wants an answer in revolutions per minute, so the conversion is: 13.7

rad 60 sec 1 rev · · = 130.8 rpm sec 1 min 2π rad

November 4, 2011

31-6 A capacitance C and an inductance L are operated at the same angular frequency. (a) At what angular frequency will they have the same reactance? (b) If L = 4.80 mH and C = 3.70µF, what is the numerical value of the angular frequency in part (a)? (c) What is the reactance of each element? (a) Equate the reactances and solve for ω: 1 = ωL ωC

1 = ω 2 LC

⇒

⇒

1 ω=√ . LC

(b) Substituting numbers, ω=p

1 (4.80 × 10−3 )(3.70 × 10−6 )

31-14 You have a 180 Ω resistor and a 0.430 H inductor. Suppose you take the resistor and inductor and make a series circuit with a voltage source that has a voltage amplitude of 28.0 V and an angular frequency of 250 rad/s. (a) What is the impedance of the circuit? (b) What is the current amplitude? (c) What is the voltage amplitude across the resistor? (d) What is the voltage amplitudes across the inductor? (e) What is the phase angle φ of the source voltage with respect to the current? (f) Does the source voltage lag or lead the current? (g) Construct a phasor diagram. (a) The impedance relates the peak values of the voltage and current. We can find it by drawing the phasor diagram: VL = ωLI

= 7.50 × 103 rad/s.

V = ZI

(c) Again substituting numbers, 1 = ωL = 36.0 Ω. ωC 31-10 You want the current amplitude through an inductor with an inductance of 5.00 mH (part of the circuitry for a radio receiver) to be 2.10 mA when a sinusoidal voltage with an amplitude of 12.0 V is applied across the inductor. (a) What frequency is required? (a) The frequency f is related to angular frequency ω by the equation ω = 2πf . The amplitude of the voltage across the inductor is VL = ILω = IL2πf . Thus the frequency is 12.0V VL = f= −3 2πIL 2π(2.10 × 10 A)(5.00 × 10−3 H) = 182 kHz

φ

VR = RI

The diagram is simpler since there is no capacitor. The voltage V of the power supply, shown by the dashed line, is p p V = (ωLI)2 + (RI)2 = (ωL)2 + R2 I = ZI. Therefore the impedance Z is given by p Z = (180 Ω)2 + (250 rad/s × 0.430 H)2 = 210 Ω.

(b) We’ve already written the relation between V and I, so I=

28.0 V V = = 0.134 A = 134 mA. Z 210 Ω

(c) To find the voltage amplitude across the resistor we use Ohm’s law and the current from part (b): VR = IR = (0.134 A)(200 Ω) = 24.0 V (d) The voltage amplitude across the inductor is the inductive reactance XL times the current: VL = ωLI = (250 rad/s)(0.430 H)(.134 A) = 14.4 V (e) Using the values of VL and VR just calculated, we can easily find the phase angle from the diagram above: φ = arctan

14.4V VL = arctan = 30.8◦ . VR 24.0V

(f) From the phasor diagram above it is clear that the power supply voltage V leads the current. (Remember the current is in phase with VR ). (g) The diagram is shown above.

Physics 21 Fall, 2011

Solution to HW-20

31-21 You have a 207 Ω resistor, a 0.408 H inductor, a 4.99 µF capacitor, and a variable-frequency ac source with an amplitude of 2.93 V. You connect all four elements together to form a series circuit. (a) At what frequency will the current in the circuit be greatest? (b) What will be the current amplitude at this frequency? (c) What will be the current amplitude at an angular frequency of 410 rad/s? (d) At this frequency, will the source voltage lead or lag the current?

31-23 In an LRC series circuit, the rms voltage across the resistor is 30.9 V, across the capacitor is 89.1 V, and across the inductor is 50.2 V. What is the rms voltage of the source? Since this is a series circuit (there is only one loop), the voltage supplied by the source will be equal to the sum of the instantaneous voltages across all the other circuit elements. However we cannot just add the rms voltages together since they are all not in phase with each other. We can use a phasor diagram to add the voltages together vectorially.

(a) We need to look at the relationship V = IZ. Since V is fixed, in order to maximize I we need to minimize Z. Z depends on ω via XL and XC : q 2 Z = R2 + (XL − XC ) Clearly, Z is minimized when XL − XC = 0. Using the definitions of XL and XC , 1 = ωL ωC 1 ω2 = LC 1 1 = 701 rad/s ω=√ =p LC (0.408 H) (4.99 µF) (b) When XL − XC = 0, Z = R, so I=

2.93 V V = = 14.2 × 10−3 A Z 207 Ω

From this we see we can use the Pythagorean theorem to determine the total voltage across all three circuit elements. p V = (VR )2 + (VL − VC )2 p = (30.9 V)2 + (50.2 V − 89.1 V)2 = 49.7 V V is the “vector sum” of the rms voltages across all the circuit elements, so the rms source voltage is V0 = 49.7 V.

(c) Now, XL − XC = (410 rad/s) (0.408 H) −

1 (410 rad/s) (4.99 µF)

= − 322 Ω I=

V =q Z

2.93 V 2

(207 Ω) + (−322 Ω)

2

= 7.66 × 10−3 A

(d) Since our calculation of XL − XC in part (c) came out negative, the circuit is mostly capacitive. This means that voltage lags current (or, equivalently, current leads voltage) at the power supply.

November 4, 2011

31-28 An L-R-C series circuit is connected to a 120 Hz ac source that has Vrms = 89.0 V. The circuit has a resistance of 74.0 Ω and an impedance at this frequency of 120 Ω. What average power is delivered to the circuit by the source? The average power in an ac circuit is given by Pav = Vrms Irms cos φ In an L-R-C circuit, the root-mean-square current is given by Vrms 89.0 V Irms = = = 0.742 A Z 120 Ω and the power factor is given by cos φ =

74.0 Ω R = = 0.617 Z 120 Ω

Thus, we can solve the average power as Pav = Vrms Irms cos φ = (89.0 V)(0.742 A)(0.617) = 40.7 W 31-59 In an LRC series circuit the magnitude of the phase angle is 50.3◦ with the source voltage lagging the current. The reactance of the capacitor is 348 Ω, and the resistor resistance is 182 Ω. The average power delivered by the source is 135 W. (a) Find the reactance of the inductor. (b) Find the rms current. (c) Find the rms voltage of the source.

ωLI

31-37 A transformer connected to a 120 V ac line is to supply 12.0 Vrms to a portable electronic device. The load resistance in the secondary is 4.40 Ω. (a) What should the ratio of primary to secondary turns of the transformer be? (b) What rms current must the secondary supply? (c) What average power is delivered to the load? (d) What resistance connected directly across the source line (which has a voltage of 120 V) would draw the same power as the transformer? (a) The ratio of the number of primary to secondary turns is equal to the ratio of the voltages of the primary to the secondary part of the transformer. So, Np Vp 120 V = = = 10. Ns Vs 12 V (b) We can determine the rms current supplied since we know the load resistance and the rms voltage supplied by the transformer. Irms =

12 V Vrms = = 2.73 A R 4.4 Ω

(c) We calculate the average power delivered to the load by using: Pav = Irms Vrms = (2.73 A)(12 V) = 32.7 W (d) We want the power on both ends of the transformer to be the same, so we can calculate the resistance that would draw this power since we know: P = IV =

RI = 182 I

V2 R

Solving for R and substituting the power found in part (c) and the source voltage we find:

φ = -50.3o V

R=

I = 348 I ωC (a) From the diagram we see that tan(−50.3◦ ) =

ωLI − 348I ωL − 348 = . 182I 182

We can easily solve for the reactance of the inductor: ωL = 129 Ω. (b) The rms current is related to the average power by p p Irms = Pav /R = 135 W/182 Ω = 0.861 A (c) Now use Pav = 135 W = Irms Vrms cos φ. We know everything in this equation except Vrms , so we solve to get 135 W = 245 V Vrms = (0.861 A) cos(−50.3◦ )

(120 V)2 V2 = = 440 Ω P 32.7 W

Physics 21 Fall, 2011

Solution to HW-22

K20-7 The wave speed on a string under tension is 170 m/s. What is the speed if the tension is doubled? From Eq. (3) in the waves handout, v = T /ρ, so if the tension T is doubled, v increases by

√ 2.

K20-11 A wave travels with speed 180 m/s. Its wave number is 1.40 rad/m. What is its wavelength? What is its frequency?

k = 2π/λ ⇒ λ = 2π/k = 2π/1.40 = 4.49 m v = f λ ⇒ f = v/λ = 180/4.49 = 40.1 Hz K20-42 This is a snapshot graph at t = 0 of a 5.0 Hz wave traveling to the left. What are the speed and phase constant of the wave?

K20-52 Earthquakes are essentially sound waves traveling through the earth. They are called seismic waves. Because the earth is solid, it can support both longitudinal and transverse seismic waves. These travel at different speeds. The speed of longitudinal waves, called P waves, is 8000 m/s. Transverse waves, called S waves, travel at a slower 4500 m/s. A seismograph records the two waves from a distant earthquake. If the S wave arrives 2.0 min after the P wave, how far away was the earthquake? You can assume that the waves travel in straight lines, although actual seismic waves follow more complex routes. Let d be the distance from the epicenter of the earthquake to the detector. Then the travel time for the P wave is tP = d/vP and the travel time for the S wave is tS = d/vS . Then tS − tP = ∆t is the delay time for the S wave, and d d − = ∆t vS vP 1 1 d − = ∆t vS vP vP − v S = ∆t d vS vP vS v P ∆t d= vP − v S Substituting numbers (2.0 min = 120 s) gives d=

4500 × 8000 × 120 = 1230 km. 8000 − 4500

15-20 A piano wire with mass 2.90 g and length 75.0 cm is stretched with a tension of 30.0 N. A wave with frequency 110 Hz and amplitude 1.50 mm travels along the wire. (a) Calculate the average power carried by the wave. (b) What happens to the average power if the wave amplitude is halved?

D(x, t) = A sin(kx + ωt + φ). From the graph, λ = 2.0 m, and f = 5.0 Hz is given. Hence v = λf = 10 m/s. We already know the graph is at t = 0. At x = 0, the amplitude is abut 0.5. Since the maximum amplitude on the graph is 1, we have ◦

0.5 = 1.0 sin(k0 + ω0 + φ) = sin φ ⇒ φ = 30 .

(a) The average power carried by the wave is given by the formula P = 12 ρA2 ω 2 v, where ρ is the mass per unit length, and A is the wave’s amplitude. We need to calculate the velocity of the wave with v = T /ρ, where T is the tension (the equation sheet uses µ for the mass per unit length): 30.0 v = T /ρ = = 88.1 m/s. (2.9 × 10−3 ) /0.75 Using ρ = m/L we can calculate the average power P =

1 .0029 (.0015)2 (2π × 110)2 (88.1) = 0.183 W. 2 .75

(b) If the wave amplitude is halved, then the average power 2 will change by a factor of 12 = 14 because power depends on the square of the amplitude. 0.183 ×

1 = .0458 W = 4.58 × 10−2 W. 4

November 11, 2011

15-24 A fellow student of mathematical bent tells you that the wave function of a traveling wave on a thin rope is D(x, t) = (2.33 mm) cos[(6.98 rad/m)x + (742 rad/s)t]. Being more practical-minded, you measure the rope to have a length of 1.35 m and a mass of 3.38 g. Determine (a) the amplitude, (b) the frequency, (c) the wavelength, (d) the wave speed, (e) the direction the wave is traveling, (f) the tension in the rope, and (g) the average power transmitted by the wave. (a) Comparing the above equation with the general form given in class, D(x, t) = A sin(kx − ωt + φ), one sees that the amplitude is A = 2.33 mm. (b) the frequency is given by f=

742 rad/s ω = = 118 Hz 2π 2π

(c) The wavelength is given by λ=

2π 2π = = 0.90 m. k 6.98 rad/m

(d) The wave speed is v=

ω 742 rad/s = = 106 m/s. k 6.98 rad/m

(e) The wave travels in the −x direction since the argument has the form of kx + ωt. (f) We get the tension T from v = T /ρ ⇒ T = ρv 2 The linear mass density ρ is ρ=

3.38 × 10−3 kg mass = = 2.504 × 10−3 kg/m. length 1.35 m

so T = ρv 2 = (2.504 × 10−3 kg/m)(106.3 m/s)2 = 28.3 N (g) The average power transmitted by the wave is given by: 1 2 2 1 1 ρA ω v = ρA2 ω 2 T /ρ = ρT A2 ω 2 2 2 2 1 = (2.50 × 10−3 )(28.3) (2.33 × 10−3 )2 (742)2 2 = 0.39 W

P =

Physics 21 Fall, 2011

Solution to HW-23

K 20-53 One way to monitor global warming is to measure the average temperature of the ocean. Researchers are doing this by measuring the time it takes sound pulses to travel underwater over large distances. At a depth of 1000 m, where ocean temperatures hold steady near 4◦ C, the average sound speed is 1480 m/s. It’s known from laboratory measurements that the sound speed increases 4.0 m/s for every 1.0◦ C increase in temperature. In one experiment, where sounds generated near California are detected in the South Pacific, the sound waves travel 7700 km. If the smallest time change that can be reliably detected is 1.0 s, what is the smallest change in average temperature that can be measured? Let D be the distance the waves travel and v be their velocity. Then their travel time is t = D/v. D dt D = − 2 ⇒ ∆t ∼ 2 ∆v, dv v v where we dropped the minus sign. We can relate the changes in velocity and temperature: ∆v ∼ 4.0 m/s, ∆T

so ∆v ∼ (4.0 m/s)∆T.

Substituting the expression for ∆v into the expression above for ∆t, we obtain ∆t ∼

YF 16-75 The sound source for a ship’s sonar system operates at a frequency f = 23.0 kHz. The speed of sound in water (assumed to be uniform at 20◦ C) is 1482 m/s. What is the wavelength of the waves emitted by the source? We know the velocity of a wave is given by: v = λf We know the frequency of the speed of sound as it propagates through water so we can determine the wavelength emitted. λ=

1482 ms v = = 0.0644 m f 23000 Hz

16-38 Two guitarists attempt to play the same note of wavelength 6.50 cm at the same time, but one of the instruments is slightly out of tune and plays a note of wavelength 6.51 cm instead. What is the frequency of the beat these musicians hear when they play together? Beats are heard when two tones with slightly different frequencies fa and fb are sounded together. fbeat = fa − fb . We also need to know that the speed of a wave is equal to the product of wavelength and frequency, v = λf . Thus, 1 1 1 1 − fbeat = v − = 344 m/s λa λb 6.50 cm 6.51 cm = 8.0Hz

D v2 ∆t. (4.0 m/s)∆T. ⇒ ∆T ∼ 2 v D(4.0 m/s)

Substituting ∆t = 1 s, v = 1480 m/s, and D = 7700 km gives ∆T ∼ 7.11 × 10−2 ◦ C. K 21-40 A violinist places her finger so that the vibrating section of a 1.10 g/m string has a length of 30.0 cm, then she draws her bow across it. A listener nearby hears a note with a wavelength of 60.0 cm. Take the speed of sound in air to be 343 m/s. What is the tension in the string? The key point is that the sound in air will have the same frequency as the vibrating violin string. Since in general λf = v, we have f=

343 m/s vair = 572 Hz = λair 0.6 m

Substituting numbers, T = (1.10 × 10−3 kg/m)(2 × 0.3 m × 572 Hz)2 = 130 N.

November 18, 2011

3

YF 16-67 A platinum wire (density ρ = 21.4 g/cm ) has diameter d = 215 µm and length L = 0.500 m. One end of the wire is attached to the ceiling, while a mass m = 440 g is attached to the other end so that the wire hangs vertically under tension. If a vibrating tuning fork of just the right frequency is held next to the wire, the wire begins to vibrate as well. What tuning-fork frequencies will cause this to happen? You may assume that the bottom end of the wire (to which the mass is attached) is essentially stationary, and that the tension in the wire is essentially constant along its length.

K 20-50 One cue your hearing system uses to localize a sound (i.e., to tell where a sound is coming from) is the slight difference in the arrival times of the sound at your ears. Your ears are spaced approximately 20 cm apart. Consider a sound source 5.0 m from the center of your head along a line 45◦ to your right. What is the difference in arrival times? Give your answer in microseconds.

(a) The problem here is to find the vibrational frequencies of a stretched wire held fixed at both ends. The wire is under tension T = mg because of the gravitational force acting on the weight. Both ends of the wire are fixed: one end is attached to the ceiling, and a heavy weight is at the other end. The fundamental mode will have a wavelength λ = 2L and a frequency f that satisfies λf = v, where v = T /µ, and µ is the mass per unit length of the wire. We find µ by dividing the total mass of the wire by its length: µ=

volume × density (πd2 /4)Lρ πd2 ρ mass = = = . L L L 4

Putting all these pieces together, we have T /µ 4mg/(πd2 ρ) v 1 mg f= = = = λ 2L 2L Ld πρ (0.440 kg)(9.80 m/s2 ) 1 = (0.5 m)(0.000215 m) π 21400 kg/m3 = 74.5 Hz. Any positive integer multiple of this frequency is a vibrational frequency of the wire.

To convert to µs, multiply the number of seconds by 106 . The result is 412 µs.

Physics 21 Fall, 2011

Solution to HW-24

16-33 Two loudspeakers, A and B, are driven by the same amplifier and emit sinusoidal waves in phase. Speaker B is 2.00 m to the right of speaker A. Consider point Q along the extension of the line connecting the speakers, 1.00 m to the right of speaker B. Both speakers emit sound waves that travel directly from the speaker to point Q. (a) What is the lowest frequency for which constructive interference occurs at point Q? (b) What is the lowest frequency for which destructive interference occurs at point Q?

16-70 Two identical loudspeakers driven by the same amplifier are located at points A and B, 2.00 m apart. The frequency is 784 Hz and the speed of sound in air is 344 m/s. A small microphone is moved out from point B along a line perpendicular to the line connecting A and B (line BC in the figure). At what distances from B will there be (a) destructive interference and (b) constructive interference? (c) If the frequency is made low enough, there will be no positions along line BC at which destructive interference occurs. How low must the frequency be for this to be the case? A LA 2.0 m

B

Speaker A is 3.0 m from Q, and speaker B is 1.0 m from Q, so the path difference is 2.0 m. (a) The speakers are in phase, so for constructive interference, the path difference must be an integral number of wavelengths (nλ). We choose n = 1 to get the lowest frequency, so λ = 2.0 m. Hence λf = v ⇒ f =

343 m/s v = = 172 Hz λ 2.0 m

(b) For destructive interference the path difference must be 0.5 wavelengths. [(n+ 12 )λ gives destructive interference, and lowest frequency is from n = 0.] Hence take λ = 4.00 m and f=

v 343 m/s = = 86 Hz λ 4.0 m

LB = x

(a) There will be destructive interference when the difference of the path lengths from each speaker tothe microphone is a half integer wavelength, or LA − LB = n + 12 λ, where n is an integer. To calculate the wavelength λ, use λ=

344 m/s v = = 0.439 m. f 784 s−1

The condition for destructive interference is LA − LB = x2 + 2.02 − x = (n + 12 )λ. Now solve the equation above for x: x2 + 2.02 = x + (n + 12 )λ

x2 + 4 = x2 + (2n + 1)λx + (n + 12 )2 λ2

x=

4 − (n + 12 )2 λ2 2(n + 12 )λ

Substituting n=0, 1, 2, 3, 4, we get x0 =0.026 m, x1 =0.53 m, x2 =1.27 m, x3 =2.71 m, and x4 =9.0 m. (b) For constructive interference, LA −LB must be an integer number of wavelengths, nλ. A similar analysis then gives 4 − n2 λ2 . 2nλ Evaluating x for several n gives x0 =0.26 m, x1 =0.86 m, x2 =1.8 m, and x3 =4.3 m. (c) As the frequency decreases, the wavelength increases, and the mimimum value LA − LB = 12 λ needed for destructive interference increases. By calculating several values of LA − √ LB + x2 + 2.02 − x, you will find that the maximum value of LA − LB is 2.0 m at x = 0, and that LA − LB gets smaller and smaller as x increases. For any frequency such that 12 λ is greater than 2.0 m, there will be no positions where LA −LB is 2.0 m. Hence at the cutoff frequency fmin , 12 λ = 2 m or λ = 4 m. We solve for this frequency using x=

fmin =

344 m/s v = = 86 Hz. λ 4m

November 18, 2011

32-11 Radio station WCCO in Minneapolis broadcasts at a frequency of 830 kHz. At a point some distance from the transmitter, the magnetic field amplitude of the electromagnetic wave from WCCO is 4.92 × 10−11 T. (a) Find the wavelength. (b) Find the wave number. (c) Calculate the angular frequency. (d) Calculate the electric field amplitude. (a) For any wave, v = f λ, and for a radio wave v = c. Hence λ=

3.00 × 108 m/s c = = 361 m. f 8.30 × 105 Hz

(b) Since we now have the wavelength, λ, we can use k=

2π = 1.74 × 10−2 m−1 . λ

(c) We were given the frequency of the wave, f , so to find the angular frequency, ω, we use ω = 2πf = 5.22 × 106 rad/s. (d) For an electromagnetic wave, the amplitude of the electric field is c times the amplitude of the magnetic field: E0 = cB0 = (3.00 × 108 m/s)(4.92 × 10−11 T) = 1.48 × 10−2 V/m. 32-16 Consider each of the electric- and magnetic-field orientations. (a) What is the direction of propagation of the wave if E = E î, B = −B ˆj. (b) What is the direction of propagation of the wave if E = E ˆj, B = B î. (c) What ˆ is the direction of propagation of the wave if E = −E k, B = −B î. (d) What is the direction of propagation of the ˆ In each case, express the diwave if E = E î, B = −B k. rection of the propagation vector as a unit vector. Its three components should be entered as x, y, z, separated by commas. For example, if the wave propagates only in the −x direction, enter −1, 0, 0. The direction of propagation is the direction of E × B. By remembering cyclic order, it’s easy to evaluate the product ˆ ˆj × k ˆ = î, and k ˆ × î = ˆj. of any two unit vectors: î × ˆj = k, ˆ ˆ ˆ ˆ ˆ (a) The direction is i × (−j) = −(i × j) = −k. Enter 0, 0, −1. ˆ Enter 0, 0, −1. (b) The direction is ˆj × î = −k. ˆ × (−î) = k ˆ × î = ˆj. Enter 0, 1, 0. (c) The direction is (−k) ˆ ˆ ˆ ˆ = ˆj. Enter 0, 1, 0. (d) The direction is i × (−k) = −(i × k)

Physics 21 Fall, 2011

Solution to HW-25

32-15 We can reasonably model a 60 W incandescent lightbulb as a sphere 5.2 cm in diameter. Typically, only about 5% of the energy goes to visible light; the rest goes largely to nonvisible infrared radiation. a) What is the visible light intensity at the surface of the bulb? b) What is the amplitude of the electric field at this surface, for a sinusoidal wave with this intensity? c) What is the amplitude of the magnetic field at this surface, for a sinusoidal wave with this intensity? (a) The intensity of light is given by the power per unit area, so the visible light intensity Ivis at the surface of the light bulb is the power due to visible light divided by the surface area of the light bulb: Ivis =

Pvis 0.05Ptot (0.05)(60.0) = = = 353 W/m2 . A 4π(d/2)2 4π(0.026)2

(b) The intensity at the surface of the bulb calculated above is just the magnitude S¯ of the time averaged Poynting vector, which is related to the electric field amplitude E0 by S¯ = 12 0 cE02 . Using this expression we can solve for the amplitude of the electric field: √ √ 2S¯ 2Ivis E0 = = = 516 V/m. 0 c 0 c (c) The amplitudes of the electric and magnetic fields are related by B0 =

E0 = 1.7 µT. c

We could have found the magnetic field amplitude directly ¯ from S¯ using an alternate expression for S: √ c 2µ0 S¯ S¯ = 12 B02 =⇒ B0 = . µ0 c

32-41 A small helium-neon laser emits red visible light with a power of 3.40 mW in a beam that has a diameter of 2.00 mm. (a) What is the amplitude of the electric field of the light? (b) What is the amplitude of the magnetic field of the light? (c) What is the average energy density associated with the electric field? (d) What is the average energy density associated with the magnetic field? (e) What is the total energy contained in a 1.00 m length of the beam? (a) The amplitude of the electric field of the light is related to the time average of the Poynting vector S¯ by 1 S¯ = 0 cE02 . 2 The time average of the Poynting vector is in units of power per unit area; it can be calculated by dividing the power given by the area of the beam: .0034 S¯ = W/m2 = 1082 W/m2 . π0.0012 Solving the above equation for electric field yields √ √ 2S¯ 2 × 1082. = = 903 V/m E0 = 0 c 0 c (b) One can find the amplitude of the magnetic field in a similar manner, using another relation from the equation sheet: √ ¯ 0 1 2Sµ c 2 S¯ = B0 =⇒ B0 = = 3.01 × 10−6 T. 2 µ0 c One could also use B = E/c. (c) The energy density associated with the electric field is uelec = 12 0 E 2 . However, this expression gives the instantaneous energy density at the time when the electric field magnitude is E. To find the average energy density associated with the electric field, you must use the time-averaged value of E 2 . Because the electric field is sinusoidal, E(t) = E0 cos(ωt) and E(t)2 = E02 cos2 (ωt). The time average of cos2 ωt = 21 , so huelec i = 12 0 hE 2 i = 41 0 E02 = 1.80 × 10−6 J/m3 (d) Similarily, the average density associated with the magnetic field is 1 2 1 hB 2 i = B = 1.80 × 10−6 J/m3 . humag i = 2µ0 4µ0 0 Note that the average energy in the electric field is the same as the average energy in the magnetic field. (e) To find the total energy contained in a 1.00 m length of the beam, use the total (average) energy density multiplied by the volume of that length. The total energy density is hutot i = huelec i + humag i = 3.60 × 10−6 J/m3 The volume of a 1.00 m length of beam is V = πr2 L = π × .0012 × 1.0 = 3.14 × 10−6 m3 , and the total energy is hutot iV = 1.13 × 10−11 J.

December 2, 2011

32-18 A sinusoidal electromagnetic wave from a radio station passes perpendicularly through an open window that has area of 0.500 m2 . At the window, the electric field of the wave has an rms value 2.60 × 10−2 V/m. How much energy does this wave carry through the window during a 30.0 s commercial? The Poynting vector gives the energy per unit area per unit time carried by the electromagnetic wave. To get the energy carried through the window in 30 s, we just multiply the magnitude S¯ of the time-averaged Poynting vector by the area of the window and then by the length of the time interval. (Since the wave propagates perpendicular ∫ to the win¯ dow, the surface integral over the window is S · dA = SA.) ¯ S is given on the equation sheet in terms of the electric field amplitude by

The fact that we get a negative result means that the image is inverted. Mastering Physics only asks for the size of the image in this part, so enter the absolute value of the result. (c) The image is inverted. (d) The fact that we got a positive result in part (a) means that the image is in front of the mirror and it is a real image. (This image could be focussed on a screen.) (e) Here is the ray diagram for this situation:

S¯ = 12 0 cE02 . Since we are given Erms , it is convenient to √ replace the peak amplitude E0 by Erms , using Erms = E0 / 2. The result is 2 . S¯ = 0 cErms

The energy is then 2 ¯ E = SA∆t = 0 cErms A∆t −12 = (8.85 × 10 )(3.00 × 108 )(0.026)2 (0.500)(30.0) = 2.69 × 10−5 J.

34-5 An object 0.550 cm tall is placed 17.0 cm to the left of the vertex of a concave spherical mirror having a radius of curvature of 22.0 cm. (a) Determine the position of the image. (b) Determine the size of the image. (c) Determine the orientation of the image. (d) Determine the nature (real or virtual) of the image. (e) Make a ray diagram and bring it to recitation. Since we are dealing with mirrors, the equations we will use are 1 1 1 + = do di f

and

m=

hi di =− ho do

The focal length f is positive since the mirror is convex, and f=

R 22.0 cm = = 11.0 cm 2 2

We take do = 17.0 cm to be positive since it is in front of the mirror. We take ho = 0.550 cm to be positive since the object is always assumed to be upright. (a) 1 1 1 = − =⇒ di f do f do 11.0 cm · 17.0 cm di = = = 31.2 cm do − f 17.0 cm − 11.0 cm (b) hi = −ho

di 31.2 cm = −0.550 cm = −1.01 cm do 17.0 cm

YF 34-8 An object is a distance of 25.0 cm from the CENTER of a silvered spherical glass Christmas tree ornament which has a diameter of 5.70 cm. (a) What is the position of its image? Use the mirror equation to answer this question, but draw a ray diagram and bring it with you to recitation. Hint: be careful to determine do (the distance to the SURFACE of the mirror) correctly. (b) What is the magnification of its image?

YF 34-14 A spherical, concave, shaving mirror has a radius of curvature of 32.5 cm. (a) What is the magnification of a person’s face when it is a distance do = 11.6 cm from the center of the mirror? (b) What is the distance di to the image? (c) Is the image real or virtual? Use the mirror equation to answer this question, but draw a ray diagram and bring it with you to recitation.

Work part (b) first: (b) The focal length is half the radius of curvature, or −16.25 cm. (distances in cm.) The sign is positive for a concave mirror. 1 1 1 1 1 1 = − = − = di f do 16.25 11.6 −40.5 The image is 40.5 cm behind the mirror. (a) The magnification is m=−

di 40.5 = = 3.49. do 11.6

(c) The image is virtual.

Physics 21 Fall, 2011

Solution to HW-26

34-23 An insect 3.00 mm tall is placed 25.0 cm to the left of a thin planoconvex lens. The left surface of this lens is flat, the right surface has a radius of curvature of magnitude 13.0 cm, and the index of refraction of the lens material is 1.70. (a) Calculate the location of the image this lens forms of the insect. (b) Calculate the size of the image. (c) Is the image real or virtual? Erect or inverted?

34-28 A photographic slide is to the left of a lens. The lens projects an image of the slide onto a wall 6.00 m to the right of the slide. The image is 80.0 times the size of the slide. (a) How far is the slide from the lens? (b) Is the image erect or inverted? (c) What is the focal length of the lens? (d) Is the lens converging or diverging?

lens do

di do + di = L

(a) The image is on the opposite side of the lens, therefore, di is positive. Since we know do is also positive, we can conclude that the magnification m = −di /do will be negative. We can write two equations for do and di from the values of L and |m| given; do + di = L di = |m|do (a) Using the formula for the focal length of a lens with two curved surfaces (the Lensmakers’ Equation), with n = 1.7, R1 = ∞, and R2 = −13 cm, we have ( ) ( ) 1 1 1 1 1 1 = (n − 1) − = 0.7 − = f R1 R2 ∞ −13 18.57 cm The value of f determined can be substituted into the lens equation: 1 1 1 = + f do di

600 cm L = = 7.41 cm. |m| + 1 80 + 1

(b) In the last part we found that m was negative, which implies an inverted image. (c) We get the focal length of the lens from the lens equation: ( ) 1 1 1 1 1 1 1 = + = + = 1+ . f do di do |m|do do |m|

f= 1 −

1 18.57

1 25

= 72.2 cm

b) To calculate the size of the image, use the magnification m=−

do =

Inverting both sides of the equation, we find

The result is: 1 1 1 = + =⇒ di = 18.57 25 di

We solve these two equations by substituting the second into the first, which leads to

72.2 di =− = −2.89 do 25.0

The image height hi is hi = mho = −2.89(3.00 mm) = −8.67mm (c) The image is real because the image distance is positive. The image is inverted because the magnification is negative. (d) If the lens is reversed, then R1 = 13 cm and R2 = ∞. The focal length is therefore: ( ) 1 1 1 1 = (0.7) − = f 13 ∞ 18.57 cm The value of 1/f is exactly the same as we had before. All the results are the same.

|m| 80 do = 7.41 cm = 7.32 cm |m| + 1 80 + 1

(d) The focal length f is positive, so the lens is converging. 33-9 Light traveling in air is incident on the surface of a block of plastic at an angle of 61.3◦ to the normal and is bent so that it makes a 47.8◦ angle with the normal in the plastic. Find the speed of light in the plastic. The speed of light in the material is given by the equation v = c/n. Snell’s Law states nair sin(θair ) = nplas sin(θplas ). The index of refraction of air is 1.00. Thus the index of refraction of the plastic is nplas = sin(θair )/ sin(θplas ). Using this in the equation for the speed of the wave we obtain vplas = c/nplas = c

sin(θplas ) sin(47.8◦ ) = (3.00 × 108 ) sin(θair ) sin(61.3◦ )

= 2.53 × 108 m/s.

December 2, 2011

YF 34-33 A diverging lens with a focal length of −47.0 cm forms a virtual image 8.00 mm tall, 16.5 cm to the right of the lens. (a) Determine the position of the object. (b) Determine the size of the object. (c) Is the image erect or inverted? (d) Are the object and image on the same side or opposite sides of the lens? Use the lens equation to answer this question, then draw a ray diagram and bring it to recitation.

ray 2 do

di

F

ray 3

(a) Since the image is virtual, we can infer that it is on the same side of the lens as the object. Therefore, its distance di will be negative. We can calculate the object distance from the Lens Equation, 1 1 1 1 1 1 = − = − = do f di −47 −16.5 25.4 cm The ray diagram shows the relative positions of the image and object. (We drew the image and object on the left, since that is the way we’ve usually done it in clsss.) (b) We determine the size of the object from the magnification m = −di /do : hi = mho =⇒ ho =

m=−

di 8.69 =− = −0.0223 do 390

Then the height of the image is hi = mho = −0.0223 × 175 = −3.90 cm = −39.0 mm (The height is negative because the image is inverted.) Comparing the height of the image to the size of the film, we see that the image will not fit upright on the film. Of course, you could tilt the camera and fit the 39 √mm image along the film’s diagonal, the length of which is 242 + 362 = 43.3 mm.

ray 1

F

(b) Obtain the image size from the magnification:

hi 8.00 mm = = 12.3 mm m −(−16.5/25.4)

(c) From the ray diagram and since m > 0, the image is upright. (d) The image and object are on the same side of the lens.

34-32 A converging lens with a focal length of 12.0 cm forms a virtual image 8.00 mm tall, 17.0 cm to the right of the lens. (a) Determine the position of the object. (b) Determine the size of the object. (c) Is the image erect or inverted? (d) Are the object and image on the same side or opposite sides of the lens? (a) Since the image is a virtual image, the sign of di is negative. The position of the object can be found by using the thin lens equation: 1 1 1 1 1 = − = − =⇒ do = 7.03 cm do f di 12 cm −17 cm (b) The size of the object is related to the size of the image by the magnification, hi = mho , where m is given by m=−

di −17 cm 0.8 cm =− = 2.42 ⇒ ho = = 0.331 cm. do 7.03 cm m

(c) The sign of ho is positive; therefore it is erect. This result can also be determined by a ray diagram. (d) The object and image are on the same side of the lens. In a lens system a virtual image is produced on the same side as the object that creates the image. The situation is illustrated by the ray diagram shown below.

34-36 When a camera is focused, the lens is moved away from or toward the film. If you take a picture of your friend, who is standing 3.90 m from the lens, using a camera with a lens with an 85-mm focal length, (a) how far from the film is the lens? (b) Will the whole image of your friend, who is 175 cm tall, fit on film that is 24 × 36 mm?

lens ho hi do

di

(a) Use the Lens Equation with do = 390 cm and f = 8.5 cm: 1 1 1 1 1 1 1 1 1 + = ⇒ = − = − = do di f di f do 8.5 390 8.69 cm The distance di from the film to the lens is 8.69 cm.

The rays are numbered 1, 2, and 3 to correspond to those described on the optics handout. Notice that on the left of the lens the rays diverge. This situation occurs when the object is closer than the focal point. In order to find the image the rays must be extended back behind the lens (shown by dotted lines) in order to create a virtual image.

Physics 21 Fall, 2011

Solution to HW-27

33-42 A light ray in air strikes the right-angle prism shown in the figure (6 B = 28◦ ). This ray consists of two different wavelengths. When it emerges at face AB, it has been split into two different rays that diverge from each other by 8.50◦ . (a) Find the index of refraction of the prism for each of the two wavelengths.

inner radius, R = 5.02 mm, what is the index of refraction of the glass? (Hint: Light from the spot on the front surface is scattered in all directions by the emulsion. Some of it is then totally reflected at the back surface of the plate and returns to the front surface.) B R/2 t

C

R/2

O

R

θc θc A

The figure above shows the rays that produce the the inner radius of the halo. A ray leaves from the point source at O, is totally internally reflected (at the critical angle θc ) at A, and then reaches the inner radius of the halo at point B. We can relate θc to the index of refraction n of the glass using nair 1 = . n n Using the geometry shown in the picture we see that we can determine θc using ( ) R R/2 −1 =⇒ θc = tan tan θc = t 2t sin θc =

(a) In order to use Snell’s Law to find the index of refraction, it is necessary to determine the angle each of these light rays make with a line normal to the surface of the prism. This means working out a little geometry. The angle θ between the incident ray and the normal to the hypotenuse of the prism is θ = 90◦ − 28◦ = 62◦ Since the 12◦ angle is given, we can find the angle between the upper refracted ray and the normal:

Using R and t given we find that θc = 35◦ . The index of refraction of the glass will be n=

1 = 1.75. sin θc

One can derive a simple expression for n by direct evaluation of sin θc using the sides of the right triangle ABC: √ 1 R/2 =⇒ n = 1 + (2t/R)2 . = sin θc = √ n (R/2)2 + t2

θ1 = 90◦ − 28◦ + 12◦ = 74◦ . The angle between the normal and the other refracted ray is θ2 = θ1 + 8.5◦ = 82.5◦ Now we can write Snell’s Law for these situations as nred sin 62◦ = nair sin 74◦

and nblue sin 62◦ = nair sin 82.5◦ ,

where nred (nblue ) is the index for the red (blue) light. We aren’t told the exact colors, but we expect the light that is bent less (more) will be more to the red (blue) end of the spectrum. Now, letting nair = 1 be the index in air, we solve for each index:

YF 34-57 A telescope is constructed from two lenses with focal lengths of 90.0 cm and 20.0 cm, and the 90.0 cm lens is used as the objective. Both the object being viewed and the final image are at infinity. (a) Find the angular magnification for the telescope. (b) Find the ABSOLUTE VALUE of the height of the image formed by the objective of a building 60.0 m tall, 3.00 km away. (c) What is the ABSOLUTE VALUE of the angular size of the final image as viewed by an eye very close to the eyepiece? Give your answer in RADIANS.

sin(74◦ ) nair sin(θ1 ) = = 1.09 ◦ sin(62 ) sin(62◦ ) nair sin(θ2 ) sin(82.5◦ ) = = = 1.12 ◦ sin(62 ) sin(62◦ )

nred = nblue

33-45 Old photographic plates were made of glass with a light-sensitive emulsion on the front surface. This emulsion was somewhat transparent. When a bright point source is focused on the front of the plate, the developed photograph will show a halo around the image of the spot. If the glass plate has a thickness, t = 3.60 mm, and the halos have an

December 8, 2011

34-89 Two thin lenses with focal lengths of magnitude 15.0 cm, the first diverging and the second converging, are placed 12.00 cm apart. An object 5.00 mm tall is placed 5.00 cm to the left of the first (diverging) lens. (a) Where is the image formed by the first lens located? (b) How far from the object is the final image formed? (c) Is the final image real or virtual? (d) What is the height of the final image? (e) Is the final image erect or inverted?

34-91 An object to the left of a lens is imaged by the lens on a screen 30.0 cm to the right of the lens. When the lens is moved 4.00 cm to the right, the screen must be moved 4.00 cm to the left to refocus the image. Determine the focal length of the lens. For the first set up, the image distance is di = 30.0 cm and let the object distance be the variable do .

(a) In a two lens system, the image from the first lens, serves as the object of the second lens. The location of the image formed by the first lens can be found using do = 5 cm, f = −15 cm, and solving the lens equation for di . 1 1 1 = + f do di Then di = −3.75. That means it the image is be 3.75 cm to the left of the lens, on the same side as object. It is an upright, virtual image. We know that it is upright because the magnification is a positive number. We know that it is virtual because di is a negative number. A ray diagram confirms this.

(b) To find the distance from the final image to the object, we need to determine the location of the final image. The image from the first lens becomes the object for the second lens. We need to determine do : because the image from the first lens is 3.75 cm to the left of the first lens, that means it is 12 cm + 3.75 cm = 15.75 cm from the second lens. Using do = 15.75 cm and f = +15 cm, solve for di = 315 cm. This means that the final image is 315 cm to the right of the second lens. So the distance between the object and final image is 5 cm + 12 cm + 315 cm = 332 cm.

Thus, the equation for the focal length is 1 1 1 1 1 = . + = + f do di do 30.0 cm

In the second set up the object distance is increased by 4.00 cm, thus d0o = do + 4.00 cm. The image distance is decreased by 8.00 cm (4.00 cm from the lens moving to the right and another 4.00 cm for the screen moving to the left). Therefore the new image distance is d0i = di − 8.00 cm = 30.0 cm − 8.00 cm = 22.0 cm.

These values can be used in another equation for the focal length 1 1 1 1 1 = 0 + 0 = + . f do di do + 4.00 cm 22.0 cm

(c) The final image is real, because we get a positive image distance (di ) for the second lens, and we can tell from the ray diagram. (d) We can find the height of the final image by multiplying the magnification of each lens. ][ ] [ 315 −3.75 − 5.00 mm = −75 mm hi = m1 m2 ho = − 5 15.75 Mastering Physics wants height (absolute value of hi ) in units of cm, so enter 7.5. (e) The final image is inverted, because the total magnification is negative, and the ray diagram confirms this.

(1)

(2)

We can set the expressions for 1/f given by Eqs. (1) and (2) equal to each other and use the quadratic equation to solve for do : 1 1 1 1 + = + do 30.0 cm do + 4.00 cm 22.0 cm 0 = d2o + 4do − 330 do = 16.3 cm. The quadratic equation for do has two roots, but do must be positive because the object is on the side of the lens from which the light comes. Therefore we used the plus sign in the quadratic formula to ensure do > 0. Now do and the original di can be used to solve for the focal length f : 1 1 1 1 1 = + = + ⇒ f = 10.6 cm. f do di 16.3 cm 30.0 cm

35-8 Young’s experiment is performed with light from excited helium atoms (λ = 502 nm). Fringes are measured carefully on a screen 1.20 m away from the double slit, and the center of the twentieth fringe (not counting the central bright fringe) is found to be 10.6 mm from the center of the central bright fringe. What is the separation of the two slits?

35-12 Coherent light with wavelength 400 nm passes through two very narrow slits that are separated by d = 0.200 mm and the interference pattern is observed on a screen L = 4.00 m from the slits. (a) What is the width of the central interference maximum? (b) What is the width of the first-order bright fringe? (a) The position x = x0 of the first dark spot is given in terms of the angle it makes with with central bright spot (at x = 0) by d sin θ ≈ d

x0 Lλ = (m + 12 )λ =⇒ x0 = , L 2d

where we invoked small-angle approximations and set m = 0 since we are interested in the central peak. We have x0 =

(4.00 m)(400 × 10−9 m) = 0.004 m = 4 mm 2(2 × 10−4 m)

The width of the central interference maximum will be double the distance from the bright central spot to the first dark spot, which is 2x0 = 8.0 mm. (b) The width of the first order bright fringe will be the distance between the first and second dark spots. We already know the position of the first dark spot. All we must do now is calculate the position of the next dark spot (at x1 , for m = 1) and find the difference (x1 − x0 ): d sin θ ≈ d

3Lλ x1 = (1 + 12 )λ =⇒ x1 = , L 2d

Clearly x1 = 3x0 = 12 mm, so the width of the first order bright fringe is x1 − x0 = 12 mm − 4 mm = 8 mm.

Physics 21 Fall, 2011

Solution to HW-28

33-31 Unpolarized light of intensity 26.0 W/cm2 is incident on two polarizing filters. The axis of the first filter is at an angle of 24.8◦ counterclockwise from the vertical (viewed in the direction the light is traveling) and the axis of the second filter is at 65.0◦ counterclockwise from the vertical. What is the intensity of the light after it has passed through the second polarizer? When the unpolarized light (intensity I0 ) passes through the first polarizer, its intensity decreases by half and it will be polarized in the same direction as the axis of the first polarizer. Therefore the intensity after the first polarizer is I1 = I0 /2 = 13 W/cm2 . The light is now polarized, so when it passes through the second polarizer its intensity decreases by another factor of cos2 φ, according to Malus’ Law: I2 = I1 cos2 (φ), where I2 is the intensity after the second polarizer, and φ is the angle between the polarization of the light and the axis of the polarizer. Since the axis of the first polarizer is at 24.8◦ , and the axis of the second polarizer is at 65.0◦ in the same direction, the angle between them is the difference, or 40.2◦ . Then I2 = (13 W/cm2 ) cos2 (40.2◦ ) = 7.58 W/cm2 . 34-58 Saturn is viewed through the Lick Observatory refracting telescope (objective focal length 18 m). If the diameter of the image of Saturn produced by the objective is 1.7 mm, what angle does Saturn subtend from when viewed from earth?

We now have all the relationships needed to answer the question. We can find θ by using the right triangle formed by the center of the objective lens, the base of the image, and the tip of the image. hi 1.7 mm = = 9.44 × 10−5 fo 18 m ( ) θ = tan−1 9.44 × 10−5 = 9.44 × 10−5 rad tan θ =

Note that we could have used the small angle approximation, tan θ ≈ θ. Mastering Physics is expecting an answer in degrees and to two significant figures, so we enter 5.4 × 10−3 . 35-21 Coherent light with wavelength 450 nm passes through narrow slits with a separation of 0.350 mm. At a distance from the slits which is large compared to their separation, what is the phase difference (in radians) in the light from the two slits at an angle of 22.6◦ from the centerline? Since the light source is coherent, the light that is emitted from each slit is in phase with each other. When the light waves arrive at the same point far from the slits, they will be out of phase by an amount proportional to the difference in their path lengths, (r2 − r1 ). We call the phase difference between these arriving waves φ. For instance, when the path difference is one wavelength, one wave has gone through one complete cycle more than the other, and φ = 2π radians. When the path difference is λ/2, φ = π radians, and so on. We can express this proportionality in the equation φ = 2π

r2 − r1 λ

If the point where the waves meet is far from the slits in comparison to their separation d, the path difference is given by r2 − r1 = d sin θ where θ is the angle measured from the centerline. Combining these equations gives φ=

2πd 2π (0.350 mm) sin θ = sin 22.6◦ = 1878 rad λ 450 nm

YF 36-24 An interference pattern is produced by two identical parallel slits of width a and separation (between centers) d = 3a. Give the number m of the first interference maxima that will be missing in the pattern. In the figure above (taken from page 1193 in the textbook), it is shown that a refracting telescope is made of two lenses, the objective and the eyepiece. Note that in this problem, we are only dealing with the objective lens. Saturn’s average distance from the earth is 1.433 × 1012 m, which is very large compared to the other distances in this problem, so we can use do = ∞. This allows us to calculate di = f = 18 m, or as shown in the figure, the objective lens creates a real image at its focal point. Note that the angle subtended by the object is the same as the angle subtended by the image. This fact follows by considering two straight rays: one from the base of the object to the base of the image, and the other from the tip of the object to the tip of the image. Both rays pass through the center of the lens and are not bent; they define the angles subtended by the object and by the image. Both angles are marked as θ in the figure.

Diffraction minima occur when a sin θ = nλ, and interference maxima occur when d sin θ = mλ. If these coincide at the same θ, then sin θ =

mλ mλ nλ mλ m nλ = = ⇒ = ⇒n= . a d 3a a 3a 3

The lowest solution for m and n is m = 3 and n = 1.

December 9, 2011

36-9 Sound with a frequency of 1250 Hz leaves a room through a doorway with a width of 1.00 m. At what minimum angle relative to the centerline perpendicular to the doorway will someone outside the room hear no sound? Use 344 m/s for the speed of sound in air and assume that the source and listener are both far enough from the doorway for Fraunhofer diffraction to apply. You can ignore effects of reflections.

K 22-19 The opening to a cave is a tall crack 43.0 cm wide. A bat that is preparing to leave the cave emits a 31.0 kHz ultrasonic chirp. How wide is the ”sound beam” 110 m outside the cave opening? Use exact formulas; don’t make small angle approximations. Use vsound = 340 m/s.

Even though this example involves sound waves, the basic phenomenon is still diffraction from a single slit. The condition for a dark fringe, or in this case a region with no sound, is given by mλ sin θ = , a where m is the order of the region of no sound (m = ±1, ±2, ±3, ...), and a is the width of the door. The wavelength of the sound can be found from the stated frequency and speed of sound: λ=

vsound 344 m/s = = 0.2752 m. f 1250 Hz

To find the angles, we solve the first equation above for θ, without making small angle approximations: mλ a = sin−1 (1 × 0.2752) = 16.0◦

for m = 1

= sin−1 (2 × 0.2752) = 33.4◦ = sin−1 (3 × 0.2752) = 55.6◦

for m = 2 for m = 3

θ = sin−1

Had we used the small angle approximation sin−1 θ ∼ θ, and converted radians to degrees, we would have obtained 15.8◦ , 31.5◦ , and 47.3◦ , for m = 1, 2, and 3, respectively. Note that the maximum possible value of m is 3; otherwise the sine becomes greater than one. 36-26 A diffraction experiment involving two thin parallel slits yields the pattern of closely spaced bright and dark fringes shown in the figure. Only the central portion of the pattern is shown in the figure. The bright spots are equally spaced at 1.53 mm center to center (except for the missing spots) on a screen 2.40 m from the slits. The light source was a He-Ne laser producing a wavelength of 632.8 nm. (a) How far apart are the two slits? (b) How wide is each one?

36-30 If a diffraction grating produces a third-order bright spot for red light (of wavelength 700 nm) at 67.0◦ from the central maximum, at what angle will the second-order bright spot be for violet light (of wavelength 430 nm)? (a) The bright spots are caused by constructive interference. From the optics handout, the condition for constructive interference is d sin θ = mλ From the information given in the problem, it is possible to determine d, the slit spacing of the diffraction grating. Setting m = 3 leads to d=

mλ 3 × 700 nm = = 2281.4 nm sin θ sin 67.0◦

Then it is possible to find the angle for a second-order bright spot for violet light. Here m = 2, and ( ( ) ) 2 × 430 nm mλ −1 −1 = sin = 22.1◦ θ = sin d 2281.4 nm

Physics 21 Fall, 2011

Information about Exam-1

First Hour Exam: There will be an exam Wednesday, Oct. 5, 2011 at 4:10 pm in CU248, PA416 and PA466; see the class web site for room assignments. Extra-time students will start in CU248, PA416 or PA466 and at 5:10 pm will move to LL221. The exam will be closed book and closed notes; the equation sheet posted on the course web site will be included on the exam. Any physical constants and integrals you need will be given on the exam. Use of Calculators and Other Electronic Devices: You should bring a calculator. Cell phones, music players and headphones are prohibited and must not even be visible during the exam. For some questions you may need to solve three equations in three unknowns. You must solve such equations by hand and show the solutions for full credit. You may use a calculator to check your hand solution. Units: In order to receive full credit, you must include the correct units with all numerical answers. Be careful about this point, because the online homework system usually provides the units for you. Coverage: You are responsible for all the reading assignments, lectures, and homework problems up to and including R-10, HW-10, and L-10. All reading assigments throught the end of chapter 27 in the text are covered. Subject Areas: The emphasis of the exam will be the material covered in lecture or recitation or that has been on the homework. The following is a list of topics and questions you should be familiar with. This list is not necessarily complete but is representative.

• What is Gauss’s Law? Be able to use it to find the field of a spherical, planar, cylindrical, or linear charge distribution. • What is the electric field inside a conductor? • In a circuit diagram, what are the symbols for battery, capacitor, and resistance? Know how to indicate the + and − terminals of a battery. • On a microscopic level, how does a dielectric (insulator) respond to an electric field? • What is Ohm’s Law? • How much power is dissipated when a current passes through a resistor? • How does a current divide when it branches to flow through two resistors in parallel? • How do the charges arrange themselves on two capacitors in parallel? in series? What is the effective capacitance for two capacitors in series or parallel? What about resistors? • What are Kirchhoff’s Rules? Know how to write the loop and junction (node) equations that we discussed in class for any circuit. What is E · dl? Know how to evaluate it for a circuit. Know how to find the potential difference between any two points on an electric circuit. • Know how to write the loop equation for an RC circuit. What is the time constant for such a circuit? Know what the function exp(−t/RC) looks like. With what time dependence does charge build up on a capacitor? With what time dependence does a capacitor discharge?

• Know the common prefixes femto through Giga.

• Know how to verify that a given solution satisfies the loop equations for a circuit.

• Know the vector form of Coulomb’s Law. Know how to find the electric field or potential of several point charges.

• Know how to account for the energy stored in the capacitor or lost in the resistor of an RC circuit.

• What is the relation between the electric field and the electric potential? Know how to evaluate the gradient. • For a distribution of charges on a line, find λ. Find σ or ρ for two- or three- dimensional distributions. • Know how to integrate over a linear charge distribution to get the electric field or potential at an arbitrary point. • Draw the electric field lines and the equipotential lines of a point charge. Do the same for a dipole. What is an equipotential surface? • What is the electric field of an infinite sheet of charge? Use this result to find the electric field between the parallel plates of a capacitor. • What is the energy density in an electric field? How much energy is stored in a charged capacitor? • What does it mean to do a surface integral E · dA? What is the convention for the direction of dA if the surface is closed?

• What is current? • What is an electron volt? • What is the dielectric constant? What happens to the capacitance and electric field if an insulator with dielectric constant K is placed between the capacitor plates? • Be able to use the right hand rule to get the direction of magnetic fields or of cross products. Know how to evaluate cross products. • In a magnetic field: What is the force on a moving charged particle? What is the force on a current? How does a charged particle move in a magnetic field? How does a velocity selector work? • What are the total force and torque on a current loop? What is the magnetic moment?

September 28, 2011

Physics 21 Fall, 2010 1

2

Hour Exam #1 3

4

Recitation Time

5

Total

Recitation Leader

Name: October 6, 2010

This exam is closed notes and closed book. You must show enough work on each problem to convince the grader you understand how to solve the problem. Make it easy for the grader to identify your final answers to each question or part of a question. You may use a calculator, but show every number that you use to get numerical results. The penalty for arithmetic errors is small if the grader can tell what you intended to do. Give units for all final answers. There is an equation sheet on the last page. All problems count 20 points. (b) (3 pts.) Determine the currents I1 , I2 , and I3 (including the correct sign) by explicit solution of the equations you determined in part (a). You must show your work.

Problem 1. Consider the following circuit: I2

a

I3

b

I1 21 V 3Ω

7Ω 2Ω 11 V

c

d

(a) (13 pts.) Write Kirchhoff’s loop and junction (or node) equations needed to determine the currents I1 , I2 , and I3 . Use the currents and their directions specified by the arrows in the diagram. You must write these equations as defined in Physics 21. You should have three equations for three unknowns. Draw clearly and label on the diagram the loop used to determine each loop equation.

(c) (2 pts.) Use the currents you calculated in part (b) to find the potential difference Vd − Va between the corner points labelled d and a on the diagram, using the path through corner point b. Show your work. (d) (2 pts.) Repeat part (c), but take the path through corner point c. Show your work, including the potential change across the resistor and battery separately. You should get the same value for Vd − Va that you found in part (c).

Problem 2. Two parallel plate capacitors are connected to a battery and a resistor as shown in the circuit. For the first capacitor, C1 = 1.24 nF; for the other, C2 = 3.72 nF. The voltage of the battery is V0 = 12 volts, and the resistance R = 3500 Ω. S

(a) V0

C1

R

Q1

Q2

C2

V0 R

(b) (6 pts.) At time t = 0 the switch S is flipped so that the circuit appears as shown in panel (b), and the capacitors discharge through the resistor. What is the time constant of the circuit? (c) (2 pts.) What is the current that starts to flow in the circuit of panel (b) just after t = 0?

S

(b)

(a) (8 pts.) Assuming that the circuit in panel (a) has been connected for a very long time, what are the charges Q1 and Q2 on each capacitor? What is the energy stored in each capacitor? Give numerical answers.

C1

C2

(d) (4 pts.) At what time t will the total charge on the two capacitors diminish to 60% of its initial value? What fraction of the initial energy will remain at time t = t ?

Problem 3. For this problem you are to find the electric field at the point P on the y axis due to the charge distribution on the x axis. A total charge of 3Q is uniformly spread out from x = −b to x = +2b. P = (0,y,0)

(b) (3 pts.) On the diagram shown, draw the vector (r − r ) (as defined in class) from the charge dQ on the rod at x = x to the point P on the y axis. Write an expression for the vector (r − r ) that indicates its components. (c) (7 pts.) Write an expression for the electric field dE at the point P due to the element of charge dQ at the point x on the rod. The expression for dE should be in terms of Q, b, dx , x , y, and NOT dQ.

dQ -b

x'

2b

(a) (3 pts.) What is the linear charge density λ on the rod?

(d) (7 pts.) Integrate to find ONLY THE y COMPONENT of the electric field E at the point P due to the rod. Your answer should be in terms of Q, b, and y.

Problem 4. A solid sphere of radius R = 2.00 mm made of an insulating material has a uniform volume charge density of ρ = 4.80 × 10−12 C/m3 .

R

(a) (3 pts.) What is the total charge on the sphere? (b) (3 pts.) Write Gauss’s Law. (c) (7 pts.) Use Gauss’s Law to find the magnitude of the electric field at a distance r = 1.00 mm from the center of the sphere. Sketch the Gaussian surface that you use and explain your result. (d) (7 pts.) Find the magnitude of the electric field at the surface of the sphere.

Problem 5. Panel (a) shows a negative ion with mass m = 1.50 × 10−26 kg and charge Q = −2.00 × 10−19 C moving along the z axis away from eight identical point charges with q = 3.25 × 10−19 C. These eight charges are fixed in the xy plane as shown in panel (b); all eight charges are the same distance b = 0.5 nm from the origin. (a) PERSPECTIVE VIEW:

(b) VIEW LOOKING DOWN z AXIS:

z Q (v = v0 k) y

b

The charges are in the xy plane at the vertices of a regular octagon centered at the origin. All charges are the same distance (b) from the origin. y q q q b

x

q x

q q

q q

(a) (7 pts.) Write an expression for the potential at an arbitrary point on the z axis due to the fixed charge distribution in the xy plane. Give your answer in terms of q, z, b, and standard physical constants. Hint: Use symmetry. (b) (6 pts.) At the instant shown in the figure, the z coordinate of the negative ion is 1.2 nm, and its speed is v0 = 15.3 km/s. Find the potential energy, kinetic energy, and total energy of the ion. Show all work and give numerical answers in joules. (c) (7 pts.) The negative ion will slow down and reach a point z = z where its speed is zero. Show explicitly the equation that must be solved to find z , and then solve that equation for z . Assume that the particle is constrained to stay on the z axis.

Physics 21 Fall, 2010

Equation Sheet

speed of light in vacuo Gravitational constant Avogadro’s Number Boltzmann’s constant charge on electron free space permittivity free space permeability gravitational acceleration

F2 on 1 =

3.00 × 108 m/s 6.67 × 10−11 N m2 /kg2 6.02 × 1023 mol−1 1.38 × 10−23 J/K 1.60 × 10−19 C 8.85 × 10−12 C2 /(N m2 ) 4π × 10−7 T m/A 9.807 m/s2

c G NA kB e 0 µ0 g

1 q1 q2 (r1 − r2 ) 4π0 |r1 − r2 |3

λ σ ; Eplane = r 20 σ for plate 0 capacitor A A Q = CV ; C = 0 K = d d 2 Q 2 Ucap = 12 CV = 12 C Uind = 12 LI 2 ρL V = IR R= A P = IV P = I 2R 1 2π0 Q E= = 0 A

1 dQ (r − r ) 4π0 |r − r |3

E = −∇V ∂V ∂V ˆ ∂V = − î + ˆj +k ∂x ∂y ∂z V f − Vi = −

f

E · dl

i

1 Q dQ 1 ; dV = 4π0 r 4π0 |r − r | 1 uelec = 12 0 E 2 , umag = B2 2µ0 V=

Work =

R = mv⊥ /(qB)

A × B =

F · dl

E · dA =

Q 0

d dt

E · dl = −

u du = a2 + u 2

a2 + u 2

du 1 = tan−1 a2 + u 2 a

v=

T /ρ

u a

ln a2 + u

1 2

(T =tension)

v = (347.4 m/s)

T /300

v = λf = ω/k ω = 2πf

k = 2π/λ

T = 1/f

(T =period)

P = 12 ρA2 ω 2 v

2

1 ωC

RC time constant = RC LR time constant = L/R Q(t) for RLC circuit Q0 exp(−Rt/2L) cos ωt ω2 =

= sin θ cos

π 2

1 R2 − LC 4L2

± cos θ sin

E · dA

u du = a2 + u 2

XR = R, XL = ωL, XC =

solenoid B = µ0 nI solenoid L = µ0 N 2 A/l

C = 2πr C = πd A = πr2 A = 4πr2 V = 43 πr3

π 2

circumference of circle circumference of circle area of circle surface area of sphere volume of sphere

cos(a ± b) = cos a cos b ∓ sin a sin b

du √ = ln u + a2 + u2 2 2 a +u

π ) 2

long wire: B =

ξeffective = ξ1 + ξ2

= ± cos θ

B · dA

d B · dl = µ0 I + µ0 0 dt

√

sin(θ ±

µ0 I 2πR center loop: B = µ0 I/2R dQ I= I = −neAvd dt Vs Ns Is Np = , = Vp Np Ip Ns µ χm = −1 µ0 τ =µ×B µ = IA

ξi =Ci (parallel) or Ri (series):

sin(a ± b) = sin a cos b ± cos a sin b

B · dA = 0

F = qv × B; dF = Idl × B µ0 Idl × (r − r ) dB = 4π |r − r |3

1 1 1 = + ξeffective ξ1 ξ2

circ. orbit ˆj ˆ k Ay Az By Bz

î Ax Bx

6.626 × 10−34 J s 1.055 × 10−34 J s 9.11 × 10−31 kg 1.6726 × 10−27 kg 1.6749 × 10−27 kg 1.6605 × 10−27 kg 8.99 × 109 N m2 /C2

h h ¯ = h/2π me mp mn u k

ξi =Ci (series) or Ri (parallel):

Eline =

F = qE dE (at r) =

Planck’s constant Planck’s constant/(2π) electron rest mass proton rest mass neutron rest mass atomic mass unit 1/(4π0 )

a−b a+b sin a + sin b = 2 cos sin 2 2

du u = √ (a2 + u2 )3/2 a2 a2 + u 2

e

du = ln u u 2π

cos2 θ dθ = 0

∂ 2D 1 ∂ 2D = 2 2 ∂x v ∂t2 1 S= (E × B) µ0

√ c = 1/ 0 µ0

c 2 S= = B0 µ0 E0 B0 Erms Brms = = 2µ0 µ0

λ = h/p

1 2

1 un+1 n+1

du 1 = ln(a + bu) a + bu b

1 du = eau a

ln u du = u ln u − u

1 cE02 2 0

un du =

u du 1 = −√ (a2 + u2 )3/2 a2 + u 2 au

ax2 + bx + c = 0 ⇒ √ −b ± b2 − 4ac x= 2a

E×B∝v ∆x∆p > ¯ h ∼

−

2π

sin2 θ dθ = π 0

KE = p2 /(2M ) (plane wave)

p=¯ hk

(¯ h = h/2π)

eiθ = cos θ + i sin θ

E=¯ hω = hf

(de Broglie)

¯ 2 ∂ 2ψ h ∂ψ = i¯ h 2M ∂x2 ∂t

August 29, 2010

Physics 21 Fall, 2010

Solution to Hour Exam #1

The graders for the problems were: 1 Jones, 2 Faust, 3 Malenda, 4 and 5 Beels For questions about the grading, see the grader by Oct. 27.

(b) (3 pts.) Determine the currents I1 , I2 , and I3 (including the correct sign) by explicit solution of the equations you determined in part (a). You must show your work.

Problem 1. Consider the following circuit: I2

a

I3 I1

loop 1

b loop 2

21 V 3Ω

7Ω 2Ω 11 V

c

loop 3

d

(a) (13 pts.) Write Kirchhoff’s loop and junction (or node) equations needed to determine the currents I1 , I2 , and I3 . Use the currents and their directions specified by the arrows in the diagram. You must write these equations as defined in Physics 21. You should have three equations for three unknowns. Draw clearly and label on the diagram the loop used to determine each loop equation.

(c) (2 pts.) Use the currents you calculated in part (b) to find the potential difference Vd − Va between the corner points labelled d and a on the diagram, using the path through corner point b. Show your work. (d) (2 pts.) Repeat part (c), but take the path through corner point c. Show your work, including the potential change across the resistor and battery separately. You should get the same value for Vd − Va that you found in part (c).

Problem 2. Two parallel plate capacitors are connected to a battery and a resistor as shown in the circuit. For the first capacitor, C1 = 1.24 nF; for the other, C2 = 3.72 nF. The voltage of the battery is V0 = 12 volts, and the resistance R = 3500 Ω. S

(a) V0

C1

R

Q1

Q2

C2

V0 R

(b) (6 pts.) At time t = 0 the switch S is flipped so that the circuit appears as shown in panel (b), and the capacitors discharge through the resistor. What is the time constant of the circuit? (c) (2 pts.) What is the current that starts to flow in the circuit of panel (b) just after t = 0?

S

(b)

(a) (8 pts.) Assuming that the circuit in panel (a) has been connected for a very long time, what are the charges Q1 and Q2 on each capacitor? What is the energy stored in each capacitor? Give numerical answers.

C1

C2

(d) (4 pts.) At what time t will the total charge on the two capacitors diminish to 60% of its initial value? What fraction of the initial energy will remain at time t = t ?

Problem 3. For this problem you are to find the electric field at the point P on the y axis due to the charge distribution on the x axis. A total charge of 3Q is uniformly spread out from x = −b to x = +2b. P = (0,y,0)

(b) (3 pts.) On the diagram shown, draw the vector (r − r ) (as defined in class) from the charge dQ on the rod at x = x to the point P on the y axis. Write an expression for the vector (r − r ) that indicates its components. (c) (7 pts.) Write an expression for the electric field dE at the point P due to the element of charge dQ at the point x on the rod. The expression for dE should be in terms of Q, b, dx , x , y, and NOT dQ.

r - r'

dQ -b

x'

2b

(a) (3 pts.) What is the linear charge density λ on the rod?

(d) (7 pts.) Integrate to find ONLY THE y COMPONENT of the electric field E at the point P due to the rod. Your answer should be in terms of Q, b, and y.

Problem 4. A solid sphere of radius R = 2.00 mm made of an insulating material has a uniform volume charge density of ρ = 4.80 × 10−12 C/m3 .

R/2

R

(a) (3 pts.) What is the total charge on the sphere? (b) (3 pts.) Write Gauss’s Law. (c) (7 pts.) Use Gauss’s Law to find the magnitude of the electric field at a distance r = 1.00 mm from the center of the sphere. Sketch the Gaussian surface that you use and explain your result. (d) (7 pts.) Find the magnitude of the electric field at the surface of the sphere.

Problem 5. Panel (a) shows a negative ion with mass m = 1.50 × 10−26 kg and charge Q = −2.00 × 10−19 C moving along the z axis away from eight identical point charges with q = 3.25 × 10−19 C. These eight charges are fixed in the xy plane as shown in panel (b); all eight charges are the same distance b = 0.5 nm from the origin. (b) VIEW LOOKING DOWN z AXIS:

(a) PERSPECTIVE VIEW:

z Q (v = v0 k) y b

x

The charges are in the xy plane at the vertices of a regular octagon centered at the origin. All charges are the same distance (b) from the origin. y q q q b

q x

q q

q

q

(a) (7 pts.) Write an expression for the potential at an arbitrary point on the z axis due to the fixed charge distribution in the xy plane. Give your answer in terms of q, z, b, and standard physical constants. Hint: Use symmetry. (b) (6 pts.) At the instant shown in the figure, the z coordinate of the negative ion is 1.2 nm, and its speed is v0 = 15.3 km/s. Find the potential energy, kinetic energy, and total energy of the ion. Show all work and give numerical answers in joules. (c) (7 pts.) The negative ion will slow down and reach a point z = z where its speed is zero. Show explicitly the equation that must be solved to find z , and then solve that equation for z . Assume that the particle is constrained to stay on the z axis.

Physics 21 Fall, 2011

Solution to Hour Exam #1

The graders for the problems were: 1 Tupa, 2 Faust, 3 Beels, 4 Malenda, 5 Glueckstein For questions about the grading, see the grader by Oct. 26.

(b) (3 pts.) Determine the currents I1 , I2 , and I3 (including the correct sign) by explicit solution, by hand, of the equations you determined in part (a). You must show your work.

Problem 1. Consider the following circuit: I2

a 8V 7Ω loop 1

c

I3

b

I1 2Ω

12 V

6Ω loop 2

d

(a) (13 pts.) Write Kirchhoff’s loop and junction (or node) equations needed to determine the currents I1 , I2 , and I3 . Use the currents and their directions specified by the arrows in the diagram. You must write these equations as defined in Physics 21. You should have three equations for three unknowns. Draw clearly and label on the diagram the loop used to determine each loop equation.

(c) (2 pts.) Use the currents you calculated in part (b) to find the potential difference Vc −Vb between the corner points labelled c and b on the diagram, using the path through corner point a. Show your work, including the potential change across any circuit elements on this path.

(d) (2 pts.) Repeat part (c), but take the path through corner point d. Show your work, including the potential change across any circuit elements on this path. You should get the same value for Vc − Vb that you found in part (c).

Problem 2. A point charge q is located at the center of a spherical cavity of radius rcav inside an insulating, spherical charged solid:

For this problem, q = −6.1 µC, rcav = 3.3 cm, the charge density in the solid is ρ = 1.5 × 10−3 C/m3 , and you are to use Gauss’ Law to calculate the electric field inside the solid at a distance r = 5.6 cm from the center of the cavity. (a) (3 pts.) Write Gauss’ Law. (b) (3 pts.) What is the shape and location of the Gaussian surface S that you will use? Draw and label S neatly on the diagram to the left.

r q

rcav

(c) (6 pts.) What is the total charge enclosed by S?

S

(d) (6 pts.) Find the magnitude of the electric field at distance r = 5.6 cm from the point charge q. (e) (2 pts.) What is the direction of the electric field? Justify your answer.

Problem 3. A battery with V = 12 V is connected to a circuit with capacitors C1 = 3.0 mF, C2 = 3.5 mF, and C3 = 2.5 mF as shown. C2 C1 C3 V

R

(a) (2 pts.) What is the equivalent (or effective) capacitance of the three capacitors shown? (b) (12 pts.) Assume the circuit has been connected for a very long time. Find the charge on each capacitor and the potential difference across each capacitor. Identify the charges as Q1 , Q2 , and Q3 , and the potential differences as V1 , V2 , and V3 . Explain carefully the steps you take to determine your answer. (c) (6 pts.) Suppose it takes the capacitors 15 s after the battery is connected to become 99% charged. What is R?

Problem 4. For this problem you are to find the electric field at the point P on the x axis due to the charge distribution on the y axis. A total charge of Q is uniformly spread out on a thin wire of length L. The lower end of the wire is at the origin. y

(b) (4 pts.) On the diagram shown, draw the vectors r and r that correspond to the field point (P ) and the charge point (where dQ is), respectively. Give the vector (r − r ) in terms of its components. (c) (6 pts.) Write an expression for the electric field dE at the point P due to the element of charge dQ on the wire. Show how to write dQ in terms of the variable of integration.

charge Q length L

dQ (0,y',0) r' r

P = (x,0,0)

x

(a) (3 pts.) What is the linear charge density λ on the wire?

(d) (7 pts.) Integrate to find ONLY THE y COMPONENT of the electric field E at the point P due to the wire. Your answer should be in terms of Q, L, x, and physical constants.

Problem 5. A particle moves in a circular orbit in a uniform magnetic field B in the −z direction (into the page). The orbit is confined to the xy plane. The charge and mass of the particle are q = 3.20 × 10−19 C and m = 6.75 × 10−26 kg, respectively, and the magnitude of B is B = 0.500 T.

(a) (5 pts.) At some time t0 the particle is at the point shown on the diagram, and its instantaneous velocity is v = (3.42 × 104 m/s) î + (3.08 × 104 m/s) ˆj. Give the components of the magnetic force F on the particle at the time t0 . Draw an arrow on the diagram that shows the direction of F. (b) (5 pts.) Find the radius R of the circular orbit.

y x +z out of page

(c) (5 pts.) How much time does it take for the particle to make one revolution? (d) (5 pts.) Through what potential difference would the particle have to be accelerated from rest to acquire the speed it has?

Physics 21 Fall, 2011

Information about Exam 2

Second Hour Exam: There will be an exam on Wednesday, November 9, 2011 at 4:10 pm. The exam will be closed book and closed notes. Any physical constants and integrals you need will be given on the exam. The equation sheet posted on the course web site will be included on the exam. Here are the room assignments: Chandler Ullman 248 Packard 466 Packard 416

Glueckstein (all), Beels (9 a.m.) and all extra time students Malenda (all), Faust (all) Tupa (all), Beels (10 a.m.)

Use of Calculators and Other Electronic Devices: You should bring a calculator. Cell phones, music players and headphones are prohibited and must not even be visible during the exam. Units: In order to receive full credit, all numerical answers must include the correct units. Be careful about this point, because the online homework system usually provides the units for you. You should be familiar with the common prefixes femto through giga. Practice Exam and Review Session: A practice exam is available from the class web site. Some of the problems on the practice exam are taken from previous exams in Physics 21. If you can work the problems on the practice exam within the time alloted, you should be well prepared for the real exam. The solution to the practice exam will be posted on the web and also discussed in a review session Tuesday, November 8 at 7:10 pm in LL270. Other review sessions may also be held; check the class website for the schedule. Coverage: The exam will cover material starting with our initial discussion of magnetism (the beginning of Chapter 27). There is some overlap with the material on the magnetic field covered on the first exam. Specifically, the exam will cover the lectures numbered 9–18 inclusive and homeworks 10–20 inclusive, as well as all the reading assignments in Chapters 27–31 of the text, except for §31.6 on transformers. Problem 31-37 on hw20 is about transformers and won’t be on the test. The exam will emphasize the material covered in lecture, recitation or homework. The following is a list of topics and questions you should be familiar with. This list is not necessarily complete but is representative. • Know the common prefixes femto through giga.

• Know why a charged particle can exhibit circular or spiral motion in a magnetic field. • How does a velocity selector work? • What is the magnetic dipole moment of a current loop? What are the forces and torques on a current loop in a uniform magnetic field? • How does a simple dc motor work? • What is Faraday’s Law? What is Lenz’s Law? Know how to apply them. • What are the magnetic and electric flux? Know how to evaluate them. • What is hysteresis? • What is ferromagnetism? diamagnetism? netism? What is a magnetic domain?

paramag-

• What is mutual inductance? What is self inductance? Why does a solenoid exhibit self inductance? • What is Ampere’s Law (final form)? Know how to use it. What is the displacement current? • What do “transient” and “steady state” mean with respect to dc and ac circuits? • Know how to analyze the transient behavior of an LR circuit. What do the exponential functions exp(−t/τ ) and 1 − exp(−t/τ ) look like? What is τ ? • What is the energy stored in a magnetic field? • Why can there be oscillations in an LC circuit? What is the resonant frequency? Be able to track where the energy is during the oscillations. What happens if you add a resistor to the circuit? • What are Maxwell’s Equations? • Know how to use phasor diagrams. You should know the phase relations between current and voltage in the prototypical ac circuits with only R, L, or C elements besides the power supply. What does it mean if, for example, the voltage leads the current across a circuit element? What is the resonant frequency of an RLC circuit with an ac power supply? • What is the power delivered to an ac circuit? • In an ac circuit, what is the relation between peak values and rms values?

• In a magnetic field, what is the force on a moving charged particle? What is the force on a current? • Be able to use the right hand rule to get the direction of magnetic fields or of cross products. Know how to evaluate cross products. • What does the magnetic field of a current loop look like? • Be able to use the Biot-Savart Law.

November 3, 2011

Physics 21 Fall, 2010 1

2

Recitation Time

Hour Exam #2 3

4

5

Total

Recitation Leader

Name: November 10, 2010

This exam is closed notes and closed book. You must show enough work on each problem to convince the grader you understand how to solve the problem. You may use a calculator, but show every number that you use to get numerical results. The penalty for arithmetic errors is small if the grader can tell what you intended to do. Give units for all final answers. There is an equation sheet on the last page. All problems count 20 points. Problem 1. For the following circuit:

(b) (3 pts.) Evaluate the peak value of the current I if the peak voltage supplied by the power supply is 8.00 V. Show all work. (c) (3 pts.) Evaluate the phase angle φ. Does the voltage lead or lag the current?

(a) (8 pts.) If R = 22.0 Ω, L = 25.0 mH, C = 10.0 µF, and ω = 800.0 rad/s, draw a phasor diagram that is approximately to scale. Include a phasor for the ac voltage V , and give the length of each phasor. Draw a mark on your diagram to indicate the phase angle φ.

(d) (3 pts.) What is the average power delivered to the circuit by the power supply? (e) (3 pts.) What is the time interval between a maximum in voltage across the inductor and the next maximum in voltage across the power supply?

Problem 2. A small circular wire loop is inside a larger loop that is connected to the circuit as shown. The values of the circuit elements are L = 15.0 H, R = 30.0 Ω, and V0 = 12.0 V. At t = 0 the switch S is closed to complete the circuit.

(c) (5 pts.) Using the box printed below, carefully draw a plot that shows the current i(t) in the large loop as a function of time, starting at t = 0. Fill in numerical values in seconds and amperes for several tic marks on the horizontal and vertical axes to set reasonable scales for these axes. (d) (3 pts.) Is the direction of the induced current in the small circular wire loop soon after the switch is closed clockwise or counterclockwise?

(a) (5 pts.) What is the time constant of this circuit? (b) (5 pts.) What is the current in the large loop a long time after the switch has been closed?

(e) (2 pts.) Now this experiment is repeated. The large loop remains the same, but the small wire inner loop is replaced with one that has half the diameter. When the switch S is closed, would the induced emf in the wire loop be smaller or larger than it was in the first experiment? Briefly explain your answer. Draw the graph for part (c) in this box:

Problem 3. A metal bar moves to the left with constant speed v = 8.00 m/s through a uniform magnetic field of magnitude B = 1.5 T as shown in the diagram. The distance between the rails is 0.5 m. The only resistance in the circuit may be taken to be the resistance R = 24 Ω shown.

(a) (4 pts.) Give Faraday’s Law (as an equation). (b) (4 pts.) What is the magnitude of the emf induced in the circuit (before the metal bar hits the resistor)? (c) (4 pts.) Is the direction of the current induced in the circuit clockwise or counterclockwise? (d) (4 pts.) Calculate the current through the resistor. (e) (4 pts.) Because of the induced current in the circuit, the magnetic field exerts a force on the moving metal bar. Find the magnitude and direction of that force.

Problem 4. The long wire shown in the diagram lies in the xy plane and carries a current I = 0.500 A. The positive z axis points out of the page.

For this problem you are to find the contribution dB to the magnetic field at several points due to the dark segment dl of the wire centered at the origin. The dark segment has a length 3.0 mm and makes an angle of 60◦ with the x axis. (a) (6 pts.) Write a vector expression for the current element I dl. (Use the unit vectors î and ˆj.) (b) (7 pts.) Find dB at x = 0, y = 0, z = 3.0 m (c) (7 pts.) Find dB at x = 4.2 m, y = 7.3 m, z = 0

Problem 5. For the following circuit, C = 25 µF and L = 32 mH. Just before the switch is closed at time t = 0, the charge on the capacitor is Q0 = 1.25 mC.

(b) (3 pts.) Give the equation that relates the current i shown on the diagram and the charge q on the capacitor. (i gives the direction of positive current flow just after the switch is closed.) (c) (6 pts.) Write the loop equation for this circuit and convert it to a differential equation for the charge on the capacitor, q(t). Verify that the solution to the differential equation is q(t) = Q0 cos ωt. (d) (4 pts.) Calculate the value of ω for this circuit.

(a) (4 pts.) What is the total energy stored in the circuit before the switch is closed?

(e) (3 pts.) What is the energy stored in the electric field of the capacitor at an instant when the magnitude of the magnetic field in the inductor is 60% of its maximum value?

Physics 21 Fall, 2010

Equation Sheet

speed of light in vacuo Gravitational constant Avogadro’s Number Boltzmann’s constant charge on electron free space permittivity free space permeability gravitational acceleration

F2 on 1 =

3.00 × 108 m/s 6.67 × 10−11 N m2 /kg2 6.02 × 1023 mol−1 1.38 × 10−23 J/K 1.60 × 10−19 C 8.85 × 10−12 C2 /(N m2 ) 4π × 10−7 T m/A 9.807 m/s2

c G NA kB e 0 µ0 g

1 q1 q2 (r1 − r2 ) 4π0 |r1 − r2 |3

λ σ ; Eplane = r 20 σ for plate 0 capacitor A A Q = CV ; C = 0 K = d d 2 Q 2 Ucap = 12 CV = 12 C Uind = 12 LI 2 ρL V = IR R= A P = IV P = I 2R 1 2π0 Q E= = 0 A

1 dQ (r − r ) 4π0 |r − r |3

E = −∇V ∂V ∂V ˆ ∂V = − î + ˆj +k ∂x ∂y ∂z V f − Vi = −

f

E · dl

i

1 Q dQ 1 ; dV = 4π0 r 4π0 |r − r | 1 uelec = 12 0 E 2 , umag = B2 2µ0 V=

Work =

R = mv⊥ /(qB)

A × B =

F · dl

E · dA =

Q 0

d dt

E · dl = −

u du = a2 + u 2

a2 + u 2

du 1 = tan−1 a2 + u 2 a

v=

T /ρ

u a

ln a2 + u

1 2

(T =tension)

v = (347.4 m/s)

T /300

v = λf = ω/k ω = 2πf

k = 2π/λ

T = 1/f

(T =period)

P = 12 ρA2 ω 2 v

2

1 ωC

RC time constant = RC LR time constant = L/R Q(t) for RLC circuit Q0 exp(−Rt/2L) cos ωt ω2 =

= sin θ cos

π 2

1 R2 − LC 4L2

± cos θ sin

E · dA

u du = a2 + u 2

XR = R, XL = ωL, XC =

solenoid B = µ0 nI solenoid L = µ0 N 2 A/l

C = 2πr C = πd A = πr2 A = 4πr2 V = 43 πr3

π 2

circumference of circle circumference of circle area of circle surface area of sphere volume of sphere

cos(a ± b) = cos a cos b ∓ sin a sin b

du √ = ln u + a2 + u2 2 2 a +u

π ) 2

long wire: B =

ξeffective = ξ1 + ξ2

= ± cos θ

B · dA

d B · dl = µ0 I + µ0 0 dt

√

sin(θ ±

µ0 I 2πR center loop: B = µ0 I/2R dQ I= I = −neAvd dt Vs Ns Is Np = , = Vp Np Ip Ns µ χm = −1 µ0 τ =µ×B µ = IA

ξi =Ci (parallel) or Ri (series):

sin(a ± b) = sin a cos b ± cos a sin b

B · dA = 0

F = qv × B; dF = Idl × B µ0 Idl × (r − r ) dB = 4π |r − r |3

1 1 1 = + ξeffective ξ1 ξ2

circ. orbit ˆj ˆ k Ay Az By Bz

î Ax Bx

6.626 × 10−34 J s 1.055 × 10−34 J s 9.11 × 10−31 kg 1.6726 × 10−27 kg 1.6749 × 10−27 kg 1.6605 × 10−27 kg 8.99 × 109 N m2 /C2

h h ¯ = h/2π me mp mn u k

ξi =Ci (series) or Ri (parallel):

Eline =

F = qE dE (at r) =

Planck’s constant Planck’s constant/(2π) electron rest mass proton rest mass neutron rest mass atomic mass unit 1/(4π0 )

a−b a+b sin a + sin b = 2 cos sin 2 2

du u = √ (a2 + u2 )3/2 a2 a2 + u 2

e

du = ln u u 2π

cos2 θ dθ = 0

∂ 2D 1 ∂ 2D = 2 2 ∂x v ∂t2 1 S= (E × B) µ0

√ c = 1/ 0 µ0

c 2 S= = B0 µ0 E0 B0 Erms Brms = = 2µ0 µ0

λ = h/p

1 2

1 un+1 n+1

du 1 = ln(a + bu) a + bu b

1 du = eau a

ln u du = u ln u − u

1 cE02 2 0

un du =

u du 1 = −√ (a2 + u2 )3/2 a2 + u 2 au

ax2 + bx + c = 0 ⇒ √ −b ± b2 − 4ac x= 2a

E×B∝v ∆x∆p > ¯ h ∼

−

2π

sin2 θ dθ = π 0

KE = p2 /(2M ) (plane wave)

p=¯ hk

(¯ h = h/2π)

eiθ = cos θ + i sin θ

E=¯ hω = hf

(de Broglie)

¯ 2 ∂ 2ψ h ∂ψ = i¯ h 2M ∂x2 ∂t

August 29, 2010

Physics 21 Fall, 2010

Solution to Hour Exam #2

If you want to discuss the grading, you must speak with the grader by Dec. 8. 1: Beels 2: Jones 3: Faust 4: Beels 5: Glueckstein Problem 1. For the following circuit:

(b) (3 pts.) Evaluate the peak value of the current I if the peak voltage supplied by the power supply is 8.00 V. Show all work. (c) (3 pts.) Evaluate the phase angle φ. Does the voltage lead or lag the current?

(a) (8 pts.) If R = 22.0 Ω, L = 25.0 mH, C = 10.0 µF, and ω = 800.0 rad/s, draw a phasor diagram that is approximately to scale. Include a phasor for the ac voltage V , and give the length of each phasor. Draw a mark on your diagram to indicate the phase angle φ.

(d) (3 pts.) What is the average power delivered to the circuit by the power supply? (e) (3 pts.) What is the time interval between a maximum in voltage across the inductor and the next maximum in voltage across the power supply?

Problem 2. A small circular wire loop is inside a larger loop that is connected to the circuit as shown. The values of the circuit elements are L = 15.0 H, R = 30.0 Ω, and V0 = 12.0 V. At t = 0 the switch S is closed to complete the circuit.

(c) (5 pts.) Using the box printed below, carefully draw a plot that shows the current i(t) in the large loop as a function of time, starting at t = 0. Fill in numerical values in seconds and amperes for several tic marks on the horizontal and vertical axes to set reasonable scales for these axes. (d) (3 pts.) Is the direction of the induced current in the small circular wire loop soon after the switch is closed clockwise or counterclockwise?

(a) (5 pts.) What is the time constant of this circuit? (b) (5 pts.) What is the current in the large loop a long time after the switch has been closed?

(e) (2 pts.) Now this experiment is repeated. The large loop remains the same, but the small wire inner loop is replaced with one that has half the diameter. When the switch S is closed, would the induced emf in the wire loop be smaller or larger than it was in the first experiment? Briefly explain your answer. Draw the graph for part (c) in this box:

Problem 3. A metal bar moves to the left with constant speed v = 8.00 m/s through a uniform magnetic field of magnitude B = 1.5 T as shown in the diagram. The distance between the rails is 0.5 m. The only resistance in the circuit may be taken to be the resistance R = 24 Ω shown.

(a) (4 pts.) Give Faraday’s Law (as an equation). (b) (4 pts.) What is the magnitude of the emf induced in the circuit (before the metal bar hits the resistor)? (c) (4 pts.) Is the direction of the current induced in the circuit clockwise or counterclockwise? (d) (4 pts.) Calculate the current through the resistor. (e) (4 pts.) Because of the induced current in the circuit, the magnetic field exerts a force on the moving metal bar. Find the magnitude and direction of that force.

Problem 4. The long wire shown in the diagram lies in the xy plane and carries a current I = 0.500 A. The positive z axis points out of the page.

For this problem you are to find the contribution dB to the magnetic field at several points due to the dark segment dl of the wire centered at the origin. The dark segment has a length 3.0 mm and makes an angle of 60◦ with the x axis. (a) (6 pts.) Write a vector expression for the current element I dl. (Use the unit vectors î and ˆj.) (b) (7 pts.) Find dB at x = 0, y = 0, z = 3.0 m (c) (7 pts.) Find dB at x = 4.2 m, y = 7.3 m, z = 0

Problem 5. For the following circuit, C = 25 µF and L = 32 mH. Just before the switch is closed at time t = 0, the charge on the capacitor is Q0 = 1.25 mC.

(b) (3 pts.) Give the equation that relates the current i shown on the diagram and the charge q on the capacitor. (i gives the direction of positive current flow just after the switch is closed.) (c) (6 pts.) Write the loop equation for this circuit and convert it to a differential equation for the charge on the capacitor, q(t). Verify that the solution to the differential equation is q(t) = Q0 cos ωt. (d) (4 pts.) Calculate the value of ω for this circuit.

(a) (4 pts.) What is the total energy stored in the circuit before the switch is closed?

(e) (3 pts.) What is the energy stored in the electric field of the capacitor at an instant when the magnitude of the magnetic field in the inductor is 60% of its maximum value?

Physics 21 Fall, 2011

Information about Final

• What is refraction? What is the index of refraction? • What is disperson?

Final Exam: The final exam will be Thursday, December 15, 2011 from 7:10–10:10 pm. Almost everyone will be in PA 101; extra time students will be in LL221 and will start 90 minutes early, at 5:30 pm. The exam will be closed book and closed notes. The equation sheet and the handout on optics (now posted on the course web site) will be included with the exam. Any physical constants and integrals you will need will be given on the equation sheet.

• Know the sign conventions for converging and diverging lenses. Know how to use the lensmaker’s equation, including the sign conventions for the radii of curvature.

Practice Questions: Representative questions on waves and optics taken from previous exams in Physics 21 will be posted on the class web site. The two previous hour exams and the associated practice exams provide representative questions on electricity and magnetism. The solution to the practice questions will be posted on the web.

• What is total internal reflection? What is the critical angle?

Use of Calculators: You should bring a calculator. In general, however, setting up the problems and demonstrating the correct strategy for solving them are worth more than doing the arithmetic to get a numerical result. If you are asked to solve simultaneous algebraic equations, you must solve them by hand and show the solution to receive full credit. Physics 19: The Physics 19 exam will consist of selected problems from the full Physics 21 exam. Physics 19 students may take the full three hours to work the exam. Coverage: The exam will cover all the material presented this semester. About 50% of the exam will be on electricity and magnetism and about 50% on waves and optics. The emphasis will be on the material covered in lecture, recitation or homework. Some questions may come from material presented only in the lectures. There may be some shortanswer questions, that is, you might be asked for a short definition or an example. You will not be asked to solve differential equations, but you may be asked to verify that a given function satisfies a differential equation. You should be familiar with all the topics and questions listed on the study guides for Hour Exams #1 and #2, as well as with the items listed below that cover optics. The list is not necessarily complete but is representative. • What is the Poynting vector? Know how to use it to describe the transport of energy by an E&M wave. • What is the difference between geometric and physical optics? What is the essential assumption of geometric optics?

• How can you describe the path of a light ray using the principle of least time? • Why is the case of parallel light rays important for a lens or mirror?

• What makes a rainbow? • What is the mathematical form of a traveling plane wave? • Know how to determine where the sound from two speakers will interfere constructively or destructively. How does your analysis depend on whether the speakers are in phase or out of phase? • What is chromatic aberration? • What is an achromatic doublet? • Be able to give examples of and explain phenomena included under the heading of diffraction and interference. What is a diffraction grating? • Be able to sketch the interference pattern for the two slit experiment if you are given the width and separation of the slits. What are missing orders and why do they occur? • What are the approximate wavelengths of visible light? • What is Huygens’ principle? • What is the f -number of a lens? • What is the difference between a real and a virtual image? • What is the difference between discrete and continuous spectra? Be able to give an example of each. • What limits the ability of a telescope to resolve two closely spaced binary stars? What is the Rayleigh criterion?

• What is the law of reflection (for a mirror)?

• What determines the polarization of an E&M wave? What is the relation between E and B and the velocity of the wave?

• Know how to use Snell’s Law.

• What does a polarizing filter do?

• What is the paraxial approximation? • Be able to draw ray diagams for mirrors and lenses using the notes handed out. Be able to draw a ray diagram for a magnifying glass or a telescope. Be able to analyize a two-lens system algebraically.

December 9, 2011

Physics 21 Fall, 2011

Practice Questions on Optics

Some of these questions on optics have been adapted from questions on the Physics 21 final given at the end of the spring semester of 2004. Try to work them using only the equation sheet and the notes on optics that will be provided with the final. Problem 1. Consider the polarizers in this problem as “perfect.” After passing through the first polarizer, the electric field of a light wave is (in SI units) E = E0 cos[6.0 × 1015 t − 2.0 × 107 x]ˆj, with E0 = 2.0 × 104 V/m.

Problem 3. A diverging lens with a focal length f = −0.06 m is 0.14 m from an object of height 0.02 m. (a) Algebraically determine the location and the height of the image. (b) Determine the location and the height of the image using a ray diagram.

(a) What are the frequency and wavelength of this wave? (b) A second polarizer is oriented at 60◦ to the first. What is the magnitude and direction of the electric field after passing through this second polarizer? y

(c) Is the image real or virtual? Problem 4. At noon on the first day of spring, the sun is directly overhead at a point on the equator; i.e., the suns shines perpendicularly to the earth’s surface at that point. Let z be the direction straight down. Assume the electric and magnetic fields of this wave are (in SI units)

60O z

(c) A third polarizer is oriented in the z direction. What is the magnitude and direction of the electric field after passing through the third polarizer? Problem 2. Two slits that are d = 2.0 × 10−4 m apart are illuminated with monochromatic light with a wavelength of 600 nm.

L (a) What is the angular separation θ between the central fringe and the next bright fringe? (b) What is the spacing (in meters) between the bright fringes if the screen is L = 3.0 m from the slits?

z î E(z, t) = E0 cos 2π 5.0 × 10 t − 6.0 × 10−7 z 14 ˆj B(z, t) = B0 cos 2π 5.0 × 10 t − 6.0 × 10−7 14

(a) What are the frequency, wavelength and velocity of this wave? (b) If E0 = 1000 V/m, what is the average intensity (power/meter2 ) of this sunshine? Problem 5. Two stars 10 light years away are barely resolved by a 90 cm (mirror diameter) telescope. How far apart are the stars? Assume λ = 550 nm and that the resolution is limited by diffraction.

Physics 21 Fall, 2011

Solution to Practice Questions

Some of these questions on optics have been adapted from questions on the Physics 21 final given at the end of the spring semester of 2004. Try to work them using only the equation sheet and the notes on optics that will be provided with the final. Problem 1. Consider the polarizers in this problem as “perfect.” After passing through the first polarizer, the electric field of a light wave is (in SI units)

Problem 2. Two slits that are d = 2.0 × 10−4 m apart are illuminated with monochromatic light with a wavelength of 600 nm.

E = E0 cos[6.0 × 1015 t − 2.0 × 107 x]ˆj, with E0 = 2.0 × 104 V/m. (a) What are the frequency and wavelength of this wave? ω = 2πf = 6.0 × 1015 ⇒ f = 9.5 × 1014 Hz k = 2π/λ = 2.0 × 107 ⇒ λ = 3.1 × 10−7 m (b) A second polarizer is oriented at 60◦ to the first. What is the magnitude and direction of the electric field after passing through this second polarizer?

y

E0

L (a) What is the angular separation θ between the central fringe and the next bright fringe? The angular splitting (in radians) is given by

O

60 E1 E2

∆θ =

30O

z

The projection of E0 on the axis of the second polarizer is E1 = E0 cos 60◦ = 1 × 104 V/m. (c) A third polarizer is oriented in the z direction. What is the magnitude and direction of the electric field after passing through the third polarizer? The projection of E1 on the axis of the third polarizer is E2 = E1 cos 30◦ = 8.7 × 103 V/m.

600 × 10−9 λ = = 0.003 d 2 × 10−4

Note that this angular splitting is small enough that the small angle approximations used are justified. (b) What is the spacing (in meters) between the bright fringes if the screen is L = 3.0 m from the slits? The spacing in meters is L∆θ = 3.0 m × 0.003 = 0.009 m = 9 mm.

Problem 3. A diverging lens with a focal length f = −0.06 m is 0.14 m from an object of height 0.02 m. (a) Algebraically determine the location and the height of the image. Let’s work the problem in cm. We have do = 14 and f = −6. The lens equation gives 1 1 1 1 1 1 − = = − = di f do −6 14 −4.2 So the image is 4.2 cm in front of the lens. The height is hi =

4.2 × 2 = 0.6 cm 14

(b) Determine the location and the height of the image using a ray diagram.

Problem 4. At noon on the first day of spring, the sun is directly overhead at a point on the equator; i.e., the suns shines perpendicularly to the earth’s surface at that point. Let z be the direction straight down. Assume the electric and magnetic fields of this wave are (in SI units)

z î E(z, t) = E0 cos 2π 5.0 × 10 t − 6.0 × 10−7 z 14 ˆj B(z, t) = B0 cos 2π 5.0 × 10 t − 6.0 × 10−7 14

(a) What are the frequency, wavelength and velocity of this wave?

ω = 2πf = 2π(5.0 × 1014 ) ⇒ f = 5 × 1014 Hz 2π ⇒ λ = 6 × 10−7 m k = 2π/λ = 6.0 × 10−7

1 object

F image

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F

(The diagram above is not exactly to scale, but it’s close enough to show how the ray diagram looks.) (c) Is the image real or virtual? The image cannot be focussed on a screen, so it is virtual.

(b) If E0 = 1000 V/m, what is the average intensity (power/meter2 ) of this sunshine? The expression that uses only the electric field amplitude E0 is S = 12 0 cE02 = .5(8.85 × 10−12 )(3.00 × 108 )(1000)2 = 1330 W/m2

Using the fact that B0 = E0 /c for a plane wave, one can write equivalent formulas involving just E0 or B0 that give the same result. Problem 5. Two stars 10 light years away are barely resolved by a 90 cm (mirror diameter) telescope. How far apart are the stars? Assume λ = 550 nm and that the resolution is limited by diffraction. The Rayleigh criterion gives θ = 1.22

λ 550 × 10−9 m = 1.22 = 7.46 × 10−7 rad D 0.9 m

The absolute distance is given by Lθ, where L is 10 ly. That works out to 7 × 1010 m, which is about 4 light minutes, or about half the distance from the earth to the sun.

Physics 21 Solutions

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