Crackiitjee.in.Phy.ch25

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CURRENT ELECTRICITY – ILLUSTRATIONS

What would he the interaction force between two copper spheres, each of mass 1 g, separated by the distance 1 m, if the total electronic charge in them differed from the total charge of the nuclei by one percent?

e. in

Assume spheres to be point charges: The atomic weight of copper is 63.55 g

= 6.022 × 1023 atoms

je

The number of atoms in 63.55 g

ac k

6.022  1023 N1 = 63.55

iit

 The number of atoms in 1 g of copper is

The number of protons in 1 g of copper is

cr

Q.

n1 = N1 × atomic number of copper = 29 N1  Total charge on the nuclei of 1 g of copper atoms is q = n1e = 29 N1e According to problem, charge on each sphere is

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q = q1 × 1% =

q1 100

=

29N1e 100

29e 6.022  1023 =  100 63.55

q2 40r2

e. in

F=

 29e  6.022  1023  9  10   63.55 

1

2

ac k

Here, e = 1.6 × 10–19 C

iit

=

2

je

9

Substitute the value of e, we get

Q.

cr

F = 1.74 × 1015 N

Calculate the ratio of the electrostatic to gravitational interaction forces between two electrons, between two protons. At what value of the specific charge

q of a particle would these forces become equal m

(in their absolute values) in the case of interaction of identical particles? Gravitational force of interaction is given by

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Fg =

Gm1m2 r2

(In magnitude) The force of electrostatic interaction between two point charges is given by Fe =

q1q2 40r2

e. in

(In magnitude) For proton,

ac k

m1 = m2 = m

iit

= 1.6 × 10–19 C,

je

q1 = q2 = + e

= 1.67 × 10–27 kg

Fe Fg

cr

 Ratio =

e2 40r 2 = Gm2 r2

e2 = 40m2G Here,

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G = 6.67 × 10–11 Nm2/kg2,

1  9  109 Nm2/C2 40 Putting the values, ratio for proton is

Fe  4  1042 Fg For electron,

e. in

m1 = m2 = m

= 1.6 × 10–19 C

ac k

 Ratio = 1036

iit

q1 = q2 = e

je

= 9.1 × 10–31 kg,

For being both forces are equal in magnitude,

cr

Fe = F g

q2 Gm2  2 or 40r2 r q2 or 2  40G m or

q  m

40G

…(i)

Here,

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0 = 8.85 × 10–12 C2/Nm2 Substituting the values, in equation (i)

q  0.86 × 10–10 C/kg m Two positive charges q1 and q2 are located at the points with radius vectors r1 and r2. Find a negative charge q3 and a radius vector r3 of the point at which it has to be placed for the force acting on each of the

e. in

three charges to be equal to zero. Since the electric force on q1 due to q2 is

q1q2 r1  r2  3

ac k

Similarly,

F13 

je

40 r1  r2

iit

F12 

q1q3 r1  r3  40 r1  r3

3

cr

Q.

and F23 

q2q3 r2  r3 

40 r2  r3

3

For equilibrium of q3,

F31  F32  0 or

q3q1 r3  r1  40 r3  r1

3



q3q2 r3  r2  40 r3  r2

3

0

…(i)

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For equilibrium of q3, the unit vector along. F31 should be in opposite direction of unit vector along F32 i.e.,

F31



F31

 F32 F32

But F31 || r3  r1  So, unit vector along F31 will be same as unit vector along r3  r1 

F31 F31



r3  r1 

ee .in



r3  r1

But

or

F32

r3  r2 r3  r2

ki

F32



ac

F32



F31

cr



itj

Also, F32 || r3  r2 

F32

F31

r3  r2 r  r   3 1 r3  r2 r3  r1

…(ii)

From Eq. (i) and (ii), we get

q3q1 r3  r1  40 r3  r1

3



q2q3 r3  r2  40 r3  r2

3

0

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or

q1q3 r3  r1 2

40 r3  r2 r3  r1 q1 r3  r2



q2 r3  r2

q2q3 40 r3  r2

2

r3  r2 r3  r2

2

r3  r1 r  r    3 2  r3  r1 r3  r2  

   or

2



q1 q2  r3  r2 r3  r2

e. in

or

q1 r3  r2  q2 r3  r2

or

q1 r3  r2   q2  r3  r2 

or r3



ac k

q1 r2  q2 r1

q1 r2  q2 r1 q1  q2

cr

 r3 



q1  q2 

iit

je

or

…(iii)

Similarly for equilibrium for q1,

r1 

q3 r2 q2 r3 q2  q3

…(iv)

On putting the value of r3 from equ. (iii) in Eq. (iv), we get

 r1 

q3 r2  q2 r3 q2  q3

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or r1



q2  q3 



q3 r2  q2 r3

or r1



q2  q3 



q3 r2  q2

 q r  q r  1 2 2 1   q1  q2   



or r1







q2  q3 

q3 r2  q2



=



 

q1  q2 r2  q2

q3

q2  q3





q1  q2



q1  q2





cr

or r1





ac k

q3

=

q2  q3

je



or r1

iit



e. in

 q r  q r  1 2 2 1   q  q   1 2

q1  q2 r2  q2



q1 r2  q2 r2



 q1 r2  q2 r1



After solving, we get q3 =



q1q2 q1  q2



2

and from Eq. (iii), we obtain

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r3 

q1  q2

A thin wire ring of radius r has an electric charge q. What will be the increment of the force stretching the wire if a point charge q0 is placed at the ring's centre? Because of force of electrical interaction, ring has a tendency to expand. So, a force of tension comes into play which balances the electrical

e. in

forces acted on different element of ring.

For calculation of force of tension, we consider an element l on the

je

circumference of ring.

iit

The charge on considered element is

ac k

q = ll where,  

q 2r

cr

Q.

r1 q2  r2 q1

The force of interaction between considered element on ring and charge q0 is. Fe =

=

q0 q 40r2

q0 l 40r2

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Here,

l  r

  Fe 2

or T 

q0 40r

q0 40r

iit

T 

je

2T sin

e. in

q0 40r

 Fe 

Q.

q0q 820r2

cr

T

q , 2r

ac k

Substituting the value of  

A thin half-ring of radius R = 20 cm is uniformly charged with a total charge q = 0.70 nC. Find the magnitude of the electric field strength at the curvature centre of this half-ring. Let us assume circular arc of radius R making an angle  at the centre of curvature of the arc. The arc is uniformly charged with charge q.

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An element of charge dq is considered on the arc. Axis yy' is bisector of angle . The electric field at point O due to considered element is

dq 40R 2

dq 

q d 

Also, d E = – dE cos  ˆj – dE sin  ˆi

e. in

dE =

dq ˆj  dq sin ˆi cos  40R2 40R2

=

q q ˆj  cos  d  sin  d ˆi 40R2 40R 2

ac k

iit

je

=

cr

 The component of electric field along y-axis is

q Ey = 40R 2 =

 / 2



cos  d

 / 2

q  /2 sin     / 2 40R2

or Ey =

q 40R2

   sin 2  sin 2   

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=

q  sin 2 20R  2

Also,  / 2

Ex =



dE sin 

 / 2

 / 2

dq sin  2  4  R 0  / 2

q = 40R 2



sin  d

 / 2

q  / 2  cos     / 2 40R2

itj

=

 / 2

ee .in

=

ac

ki

 Ex = 0

 Net electric field is

E2x  E2y

cr

E=

E 

2  sin 2 20R 2

(along bisector of angle ) For semi-circular ring,



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E

q  sin 2 20R 2

E 

q  sin 2 2 0R 2

=

2

q 2 0R 2 2

Substituting the values, we get

A point charge q is located at the centre of a thin ring of radius R with

je

uniformly distributed charge –q. Find the magnitude of the electric

iit

field strength vector at the point lying on the axis of the ring at a

ac k

distance x from its centre, if x >> R.

This problem is based upon superposition principle of electric field.

E  E1  E2

cr

Q.

e. in

E = 0.1 kV/m

Here,

E = resultant electric field,

E1 = electric field due to ring.

E2 = electric field due to point charge on the centre of ring. According to superposition principle E1 and E2 are physically independent to each other.

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The electric field at point P due to point charge q is

E2 

q ˆ i 40x2

(along axis of ring) Electric field at point P due to ring is



qx 2

2

40 R  x



ˆi

3 /2

e. in

E1 

 E  E1  E2



iit



  ˆi 3 /2  

je

 q 1 x  = 40  x2 R 2  x2 





ac k

3 /2 2  2  x3  q  R x  ˆi or E  3 /2 2 2 2  40  x R  x  



cr



3 /2    R2  3 3 x 1  2  x  x  q    = 3 / 2  40  x2 R 2  x2    





 q  3 3R 2 x 1   ....    2 40   2x  or E  3 /2 x2 R 2  x2





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Since, x >> R, so in binomial expansion higher power may be neglected.

E 

q 40

 3R 2x   2 i   3 /2 x2 R 2  x2





But R2 may be neglected with respect to x2.

3qR 2x E  80x5

je

A thin non-conducting ring of radius R has a linear charge density  =  0

iit

cos , where0 is a constant,  is the azimuthal angle. Find the

ac k

magnitude of the electric field strength: (a)

at the centre of the ring;

(b)

on the axis of the ring as a function of the distance x from its

cr

Q.

e. in

3qR 2 = 80x4

centre. Investigate the obtained function at x >> R.

Here, the meaning of azimuthal angle is the angle with diameter. Since,  is cosine function of . So, ring if not uniformly charged. A part of ring is positively charged and remaining part of ring is negatively charged. The distribution of charge on ring.

  0 cos 

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So, first and fourth quadrants are positively charged but second and third quadrants are negatively charged. Now we consider a small element dx = Rd  on the ring. The charge on considered element is dq = dx = 0 cos  Rd 

 Net electric field at point P is.

40 R 2  x2



3 /2

iit



 0R 2

ac k

=

je

E = E1 + E2

e. in

= 0 R cos  d

when x >> R, R2 may be neglected with respect to x2.

cr

 0R 2 E= 4 0 x3 when x >> R

 0 R 2 = 40x3 =

p 40x3

where, p =  0R2

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(a)

For electric field at the centre of ring, x = 0,

E

 0R 2



40 R 2  x2



3 /2

At centre, x = 0

 0R 2 E  40R 3

Let the ring plane coincides with y-z plane. We consider a small

iit

je

element AB (of length dl) on ring.

ac k

Here dl = RdQ

where R is the radius of ring. y = R sin 

cr

(b)

0 40R

e. in

=

and z = R cos  The electric charge on the considered element is dq = dl = 0 cos  (R d) = 0 R cos d The axis of the ring is X-axis. The electric field at point P due to considered element is

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dE 



dq rp  rA

 3

40 rp  rA

 0R cos  d  xˆi  yjˆ  zkˆ

or dE 

ˆ ˆ  zk xˆi  yj

40



3



ˆ ˆ ˆ  0R cos  d x i  R sin j  R cos k or dE  3 /2 40 x2  y2  z2



ee .in







ˆ xˆi  R sin ˆj  R cos k  0R cos  d = 3 /2 40 x2  R 2 sin2   R 2 cos2 



itj







ˆ ˆ ˆ  0R cos  d x i  R sin j  R cos k = 3 /2 40 x2  R 2

ac

ki





 0R

40 R 2  x2

cr

 dE 





3 /2

 x cos  d ˆi  R sin  cos  d ˆj   ˆ   R cos2  R cos2  d k

 dEx 

 0Rx cos  d



40 R 2  x2



3 /2

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dEy =

  0R 2 sin  cos  d



40 R 2  x2



3 /2

 0R 2 cos2  d

and dEz =



40 R 2  x2



3 /2

 Ex   dEx



2

2

40 R  x



3 /2

After integrating,

 dE

y

ki

and Ey =

0

itj

Ex = 0

 cos  d

ee .in

=

2

 0Rx

=

2

ac 

40 R 2  x2

cr

=

 0R 2

 0R 2



80 R 2  x2



3 /2

 0

sin2 d 2 2



3 /2

  cos 2    0 2  0

 Ey  0 Similarly, Ez =

 dE

z

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=

=

2

 0R 2



40 R 2  x2



 0R 2

40 R 2  x2





 cos

2

3 /2

 d

0

3 /2

ˆ  E  Ex ˆi  Ey ˆj  Ez k



 0R 2

40 R 2  x2

3 /2

ac k

For x >> R,



je

=



Ex  0, Ey  0

iit



e. in

ˆ  E  E  Ez k

R2 + x2  x2

 0R 2 40x3

cr E 

Q.

A very long straight uniformly charged thread carries a charge  per unit length. Find the magnitude and direction of the electric field strength at a point which is at a distance y from the thread and lies on the perpendicular passing through one of the thread's ends. Since we know that

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E|| =

 cos 1  cos 2  40r

E 

 sin 1  sin 2  40r

The net electric field is

E2  E||2

E=

e. in

The situation of problem is Here,

iit

je

1 = 0°

 2 

 2

cr

and r = y

ac k

Since, two parallel lines meets at infinity.

E 

   sin0  sin 40y  2 

 E|| 

=

   cos 0  cos   40y  2

 40y

The net electric field is

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E=

E2  E||2

or E =

2  40y

Also, tan  =

E 1 E||

e. in

   45 It means electric field E is directed at the angle 45° to the thread.

je

A thread carrying a uniform charge  per unit length has the

iit

conurations. Assuming a curvature, radius R to be considerably less

ac k

than the length of the thread, find the magnitude of the electric field strength at the point O.

This problem is based upon superposition principle of field for continuous distribution of charge.

cr

Q.

E  E1  E2  E3 (a)

The thread of the problem may be divided into three parts. (i)

Semi-infinite vertical thin thread.  We know that

E 

 40R

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E|| 

 40R

so, the net electric field due to this part of thread at point O is

E1  Eˆi  E|| ˆj Putting the value of E and E||, we get

E1 

 ˆ  ˆ i j 40R 40R

je

Second part:

iit

(ii)

e. in

We get

ac k

A circular arc which is making angle

 at the centre of 2

curvature O

cr

 The electric field due to this part is E2 =

=

=

q  sin 20R2 2

 R  20R

2

sin

 2

  sin 20R 2

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  90 2



 4  E2  20R  sin

=

 2 20R

e. in

In vector from,

E2  E2 cos 45ˆi  E2 sin 45ˆj

je

 1 ˆ   1 ˆ  i  j 2 20R 2 2 20R  2 

iit

or E2 

 ˆ  ˆ i j 40R 40R

ac k

or E2 

Third part:

cr

(iii)

This consists of a semi-infinite thin horizontal thread.

E 

 40R

E|| 

 40R

The electric field at point O due to this part is

E3  E|| ˆi  E ˆj

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=

 ˆ  ˆ i j 40R 40R

The net electric field at point 0 is

E  E1  E2  E3. Putting the values of E1, E2 and E3, we get

E  E2

(i)

ac k

The thread of problem may be divided into three parts:

First part:

cr

(b)

2  40R

iit

E 

e. in

 ˆ  ˆ i j 40R 40R

je

=

Semi-infinite vertical thread. The electric field at point O due to this part is

E1  (ii)

 ˆ  ˆ i j 40R 40R

Second part: This part is semicircular thread of radius R. The electric field due to circular arc is

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E2 =

  sin 20R 2

In the case of semicircular arc, =

 E2 

 20R

In vector form,

(iii)

 ˆ j 20R

Third part:

ee .in

E2 

ac

ki

itj

Semi-infinite vertical thin thread

The electric field due to this part is

cr

E3 

 ˆ  ˆ i j 40R 40R

 Net electric field due to thread at point O is

E  E1  E2  E3 Substituting the values of E1, E2 and E3, we get,

E  0.

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A very long uniformly charged thread oriented along the axis of a circle of radius R rests on its centre with one of the ends. The charge of the thread per unit length is equal to . Find the flux of the vector E across the circle area. Since the solid angle subtended by a right circular cone on its vertex is  = 2(1 – cos ), where  is semivertex angle.

e. in

Electric flux through art area ds due to a point charge q.

q dx cos  40r2

=

q ds cos  40 r2

ac k

=

iit

  E.ds

je

Electric flux  is defined as

cr

Q.

  E.ds =

q ds cos  40r2

 

q 40

where,  is solid angle subtended by area ds on the point charge.

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In similar fashion, electric flux passing through a given area due to an elementary charge dq is

d 

dq 40

where  is solid angle subtended on the elementary charge dq. We consider an element dx at distance x from centre of circle.

e. in

The electric charge on the considered element is dq = dx. (treated as point charge).

je

The solid angle subtended by circle on the considered element is  =

Here,

x

R 2  x2

cr

cos  =

ac k

iit

2(1 – cos ),

  x    2 1    R 2  x2  The electric flux passing through the circle due to considered element is

d 

=

dq 40

2 40

 1 

   dx R2  x2  x

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 Total electric flux crossing through the circle due to large thread is

   d

 = 20

1

  1  0

  dx R 2  x2  x

Here, put x = R tan dx = R sec2 d

R tan    1  R sec2  d     R sec   x 0

1

 1  sin  R sec

d

je

2

0

iit

 = 20

x 1

e. in

   20

R  x 1 x 1  tan   sec      x0   x 0 20

cr

=

ac k

x 1 x 1  R  2 =   sec d   tan  sec d 20  x  0 x0 

 

R x 1 tan   sec x  0  20

R  20

R = 20

1

x R 2  x2     R  R 0

l  R 2  l2   1   R  R 

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Thread is very long.  R >> 1

R  20

l l    R R

 R2  1  1 l2 

R2 may be neglected l2

R 20

e. in

 

l l    R R  1  

je

R 20

iit

Two point charges q and –q are separated by the distance 2l. Find the

ac k

flux of the electric field strength vector across a circle of radius R.

Electric flux through an area due to a number of point charges is given by

cr

Q.

 

1  40

n

q  i

i

i 1

The solid angle due to circle on the point charge q is  = 2 (1 – cos ) Since, charges are opposite sides of the circle. So, net flux passing through circle is  = 1 – 2

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Here,

1 

q1 40

=

q2 1  cos   40

=

q 1  cos   20

q2 40

q2 1  cos   40

=

 q 1  cos   20

cr

ac

ki

=

itj

2 

ee .in

Similarly,

  1  2 =

q 0 1  cos  

cos  =

l R 2  l2

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 

 1 

q 0

  R 2  l2  l

Q.

r unit length. The threads are separated by a distance

. Find the

maximum magnitude of the electric field strength in symmetry plane of this system located between the threads. The electric field due to a long thread (thin rod) is

 20r

ee .in

E=

The direction of electric field is along increasing r.

itj

Let both threads are placed perpendicular to the plane of paper (along z-

ac

ki

axis) at a separation of l.

cr

Horizontal components of electric field balance each other.  Net electric field at point P is E = 2E1 cos  But electric field due to one of the thread is: E1 =

 20r

E 

2 cos  20r

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=

 cos  0 r

=

 y 0 r2

E=

 y 2 0  l 2  4  y 

e. in

For maximum electric field,

je

dE 0 dy

l 2

 l   2   0 l2 l2  4 4

cr

 Emax

ac k

y=

iit

After solving, we get

= Q.

  0l

An infinitely long cylindrical surfaced of circular cross-section is uniformly charged lengthwise with the surface density  = 0 cos , where  is the polar angle of the cylindrical coordinate system whose z

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axis coincides with the axis of the given surface. Find the magnitude and direction of the electric field strength vector on the z axis. Here, since, surface charge density is given. So, the cylinder may be treated as hollow charged cylinder. The given cylinder behaves as a long bamboo. So, the strip of the cylinder alone the cylinder behaves as long thread.

e. in

The top view of cylinder is like a circular ring.

Now we consider a long strip of width dt = Rd If length of cylinder is l(l

je

).

ac k

dA = ldt = lRd

iit

The area of strip is

The charge on the strip is

cr

dq = dA

=0 coslRd

= 0 lR cos d The linear charge density of considered strip is,



=

dq l

0lR cos d l

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= 0R cos  d The electric field at point O due to considered long strip is

 20R

=

0R cos  d 20R

=

0 cos  d 20

e. in

dE =

je

 dE   dE cos  ˆ i  dE sin ˆj

 2 



dE cos 

 0

ac k

Ex =

iit

The component of electric field along x-axis due to whole cylinder is

0 = 20

2

0 = 20

2

0 = 40

2

 cos

 d

cr

2

=

0

 1  cos 2   d 2

  0

 1  cos 2 d 0

0 2 0

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The component of electric field due to whole cylinder along y-axis is  2 

Ey =



dE sin 

 0

 = 0 20

2

0 = 40

2

 cos  sin  d 0

 sin2 d  0

 Net electric field at point O is

Q.

0 2 0

cr

i.e., E =

iit

0 ˆ i 2 0

ac k

=

je

i  Ey ˆj E = Ex ˆ

e. in

0

A ball of radius R carries a positive charge whose volume density depends only on a separation r from the ball's centre 0 (1 –

r ), R

where 0 is a constant. Assuming the permittivities of the ball and the environment to be equal to unity, find: (a)

the magnitude of the electric field strength as a function of the distance r both inside and outside the ball;

(b)

the maximum intensity Emax and the corresponding distance rm.

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This problem can easily be solved by using Gauss's law.



i.e., E.dS  c



or E.dS 

qin 0

 dq

c

0

e. in

Case-I.

c

qin 0

c

qin 0

cr

E 4r2 

qin 0

ac k

 EdS cos 0 

E 

iit

 E.dS 

je

When r < R

qin 40r2

…(i)

where, qin = Total electric charge enclosed in a sphere of radius r Here, r

qin =

 4r dr 2

0

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r

=



0

0

r  2 1  R  4r dr

r 3 r 2  r = 40   r dr   dr  R 0 0 

 r3 r 4  = 40     3 4R 

e. in

40  r3 r 4  =  0  3 4R  From equation (1),

je

40  r3 r 4   4r2  3 4R 

ac k

=

qin 40r2

iit

E=

r r2    3 4R  radially outward  

0 0

E 

0r 30

cr

E 

3r   1   4R  

where r < R Case-II. When r = R at the surface of sphere

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0  R R 2  E=  0  3 4R 

 4  3

= 0R    12 

E 

0R radially outward. 120

Case-III.

e. in

When r > R

je

In this case, R

iit

r  2  1  0 0  R  4r dr

ac k

qin =

cr

R3 R 4  or qin = 40   4R  3

1 1    3 4 

or qin = 40R 3 

3 or qin = 40R 

1 12

0R 3 = 3 From equation (i), we have

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E=

qin 40r2

0R 3 = 3  40r2 0R 3 E  120r2

0r 30

3r   1   4R  

e. in

E 

where r < R

je

0R 120

iit

E=

ac k

when r = R

cr

0R 3 and E = 120r2 when r > R (b)

From expression of electric field, the electric field outside sphere decreases with increasing r. Hence, electric field will be maximum inside the sphere.

E

0r 30

3r   1  4R   

 For maximum electric field,

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dE 0 dr or 1 

3r 0 2R

r 

2R 3

=

 3  2R   1   4R  3   

ac

A system consists of a ball of radius R carrying a spherically symmetric charge and the surrounding space filled with a charge of volume

 , where  is a constant, r is the distance from the centre r

cr

Q.

 0R 9 0

 2R  0   3   30

itj

 Emax

ee .in

2R , electric field will be maximum. 3

ki

At r =

density  =

of the ball. Find the ball's charge at which the magnitude of the electric field strength vector is independent of r outside the ball. How high is this strength? The permittivities of the ball and the surrounding space are assumed to be equal to unity. The electric field on the surface of a sphere is

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E=

q 40r2

And this result is independent of distribution of charge inside the sphere. Now, we take origin at the centre of sphere. From r = 0 to r = R, space is uniformly charged (volume charge density is constant). From r = R to r , volume charge density is  

 . For convenience, r

e. in

the system may be assumed as combination of a uniformly charged sphere of radius R and a hollow sphere of inner radius R1 = R and outer radius R2

ac k

E  E1  E2

iit

fields due to both spheres.

je

. The resultant electric field at a point is calculated by vector sum of

Calculation of E1:

q0 at point P at a distance x from centre of sphere. 40x2

cr

E2 =

Here, q0 = charge on the sphere. Calculation for E2: E2 =

q where q is charge on the hollow sphere from r = R to r = x. 40x2

Analysis for q.

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We consider a hollow sphere of radius r (R < r < x) and thickness dr. The volume of considered element is dV = 4r2dr.  Electric charge on the considered element is dq =dV

 4r2dr r

e. in

=

= 4rdr

The total charge enclosed from r = R to r = x is

je

 dq

iit

q=

x

R

ac k

 rdr

= 4

cr

 x2  R 2  = 4  2  



= 2 x2  R2



 Net electric field is E = E1 + E2 (both are in the same direction). E=

q0 q  2 40x 40x2



2 x2  R 2 q0  or E = 40x2 40x2



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(on putting the value of x) or E =



q0  2 x2  R 2



2

40x

q or E =

0



 2R 2  2x2 40x2

According to problem, result should be independent of x (in the

e. in

problem, r is taken at the place of r). If q0 – 2R2 = 0

2 which is independent of x. 40

Hence,

ac k

q0 – 2R2 = 0

iit

je

Then E =

 q0 = 2R2

 2 0

cr

and E =

Q.

A space is filled up with a charge with volume density 0 er

3

where 0 and  are positive constants, r is the distance from the centre of this system. Find the magnitude of the electric field strength vector as a function r. Investigate the obtained expression for the small and large values of r, i.e. at r3 > 1. The electric field on the surface of spherical distribution of charge, is

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E= 

q 40r2

This result is applicable for uniform as well as non-uniform distribution of charge. This result does not depend upon charge outside the surface. Draw a sphere of radius r. Now, calculate the net charge enclosed by the sphere.

e. in

For this, we consider a hollow concentric sphere of radius r and thickness dr. The volume of considered element is dV = 4r2dr.

je

The electric charge in the considered element is

iit

dq = dV 3

ac k

= 0er 4r2dr 3

cr

= 40r2er dr

 Total charge enclosed by the sphere is r

q=

 4 r e 0

2

r3

dr

0

q 

E



3 40 1  er 30



q radially outward 40r2

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=

3 0 1  er   30r2 

Let us discuss the condition of problem: According to problem,

r3  1 3

er  1  r3

0r 3 0

je

=

e. in

0r3 E= 30r2

3

ck

iit

er  0 (whenr3 >>> 1)

0 30r2

cr a

E 

Q.1 Two metal balls of the same radius are located in homogeneous poorly conducting medium with resistivity (). Find the resistance of the medium between the balls provided that the separation between them is much greater than the radius of the ball. Sol. ∵ Electric resistance is R =∫ Let us consider a hemispherical shell of radius r and thickness dr whose centre is at O. The electric resistance of considered element is dR =  ∴ The resistance between sphere is

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

R=∫ R=



0

=



l a

 1  r  =  a



0

1=

1 But l >>> a, ∴ l – a  l ∴ R = )

(



0

(

)

1



Alternative method: Ohm’s law in the vector method is: ⃗ = Where = current density

e. in

At first, spheres are charged by opposite nature of charge of magnitude q. The electric field at the surface at the first ball is E =

je

Here the second ball is infinitely at large distance from first ball. The current density J =



iit

So, electric field due to second ball on the surface of first ball will be zero.

ac k

J=

Electric current I = ∫ ⃗



dS

cr

=∫ =

= J∫ 4



=j×4

=

But the potential difference between ball is =

-

=

-.

∴ Electric resistance

R=

(from Ohm’s law)

∴

R=

/

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Q.2 A metal ball of radius a is located at a distance l from an infinite ideally conducting plane. The space around the ball is filled with a homogenous poorly conducting medium with resistivity ( ). In case a ≪ 1 find: (a) The current density at the conducting plane as a function of distance r from the ball if the potential difference between the ball and the plane is equal to V; (b) The electric resistance of the medium between the ball and the plane. Sol. Potential difference between the ball is =V= V

(1)

ee .in

q=

(a ) The electric field at the point P is due to charge q is E=

∴ Net electric field at point p is E = 2E0 cos E= E=

cr ac

E=

according to Ohm’s law, ∴

from equ. (1)

ki

or or

cos

itj

or

E= J

J=

=

(b) Now we consider a hemispherical shell of thickness dx and radius x whose centre is at the centre of the ball. =∫ 

R=∫ =



0

= 1=

l a

 1  r   0





(∵ l > > a)

Q.3 Two long parallel wires are located in poorly conducting medium with resistivity ( ). The distance between the axes of the wire is equal to l, the cross section radius of each wire equals a. in the case a ≪ l find;

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(a) the current density at the point equally removed from the axes of the wire by a distance r if the potential difference between the wires is equal to V. (b) The electrical resistance of the medium per unit length of the wires. Sol. (a)

The electric field at a point P at a distance r from both wires is E= By ohms law E = J

(b)

=

ee .in

J=

The electrical field near first wire is

=

ki

J=

itj

E0 =

cr ac

I=jS=j2 a R= R=

=

(per unit length) (per unit length) =

Ans.

Q.4 The gap between plates of a parallel plate capacitor is filled with a glass of resistivity ( ) = 100 G.m. The capacitance of the capacitor equals C = 4.0 nF. Find the leakage current of the capacitor when a voltage V = 2.0 kV is applied to it. Sol. The problem is solved in following steps: Find resistance and capacity of system separately. Here

R=

Also,

C=

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∴

=

Now draw equivalent circuit In loop AGEFA,

=0

∴

q0 = CV

In loop ABDFA,

- IR + V = 0

∴

V = IR

∴

I=

=

=

ee .in

=

Here current I through equivalent resistance of the system is known as leakage current. According to problem,

C=4×

itj

 = 100 ×

m

ki

= dielectric constant for glass = 6

cr ac

Substituting the values we get,

I = 1.5 ×

A = 1.5 A

Q.5 Two conductors of arbitrary shape are embedded into an infinite homogenous poorly conducting medium with resistivity () and permittivity ( ). Find the value of a product RC for this system, where R is the resistance of the medium between the conductor, and C is the mutual capacitance of the wires in the presence of the medium. Sol. Here Ohm’s law in the vector form is applicable i.e. ⃗ =  ∵ Conductors are oppositely charged by charge q. The electric field at the surface of conductor is E = Here But

in the direction of area.

= E = J (here ⃗  ) ∴

J=



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Since, electric field and area is in same direction so, angle between area vector and current density vector should be zero. ∴

J=

Or

I=

∴ I = JS

(where s is area)

or V = IR



(Here V is potential difference between conductors) = IR

Or ∴

RC = =

=

= Ans.

e. in

RC =

je

Q. 6: A conductor with resistivity bounds on a dielectric with permittivity , at a certain point A at the conductor’s surface the electric displacement equals D, the vector D being directed away from the conductor and forming an angle a with the normal of the surface. Find the surface density of charges on the conductor at the point A and the current density in the conductor in the vicinity of the same point.

ac k

iit

Sol. ∴ Surface charge density is ∴

=D

D=

E

E=

at point

(normal component)

cr

And

=

The tangential component of electric field inside conductor is E0 = E By Ohm’s law,

E0 =

∴

J=

=

=

Q.7 The gap between the plates of a parallel plate capacitor is filled up with an inhomogenous poorly conducting medium whose conductivities varies linearly in the direction perpendicular to the plates from 1 = 1.0 pS/m to 2 = 2.0 pS/m. Each plate has an area S = 230 cm2, and the separation between the plates is d =

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2.0 mm. Find the current flowing through the capacitor due to a voltage V = 300 V. Sol.

∵

R=∫

=

and



Here  is linear function of distance x.  = mx + K when x = 0,  = 1 1 = m × 0 + K K = 1 when x = 0,  = 2 2 = md + K

e. in

∴

m=

∴

=.

/ x + 1

iit

∴

je

2 = md + 1

Now solve the problem in following ways:

ac k

Find equivalent resistance, we consider an element of thickness dx at distance x from first plate. The electric resistance of considered element is

cr

dR = 

=



=

0.

1

/

The equivalent resistance between plates is R=∫

=∫ =

(

0.

/

1

)

Now draw equivalent circuit of system : The system behaves as parallel combination of a resistor of resistor R and a capacitor of capacitance C. V = IR

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I= Putting the value of R1, we get (

I=

)

Putting the values, we get I = 5 nA Q.8. Demonstrate that the law of refraction of direct current lines at the = where 1 boundary between two conducting media has the form

e. in

and 2 are the conductivities of the media, and are the angles between the current lines and the normal of the boundary surface.

je

Sol. Since the electric charge is neither pilling up nor disappearing at a boundary, the current toward the boundary on one side must be equal to that away from it on the other side. By conservation charge,

iit

I1 = I2 ⃗⃗ . d ⃗ = ⃗⃗ . d ⃗

or

(J1 cos 1) dS = (J2 cos 2) dS

or

J1n = I2n ………………………….. (i)

cr a

ck

or

(normal component of J) J=E

By Ohm’s Law,

1 E1n = 2 E2n ……………… (ii) But there is no source of emf. ∴ Here  or

∫⃗ . d ⃗ = 0 E1n = E2n = =

………………………….. (iii)

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(∵ J1t and J2t are horizontal components of J1 and J2 respectively) From equation (i) and (iii), we get ∵

1 cot 1 = 2 cot 2

Here,

1 = 1 and 2 = 2

∴

=

∴

=

e. in

Q.9. Two cylindrical conductors with equal cross sections and different resistivities 1 and 2 are put end to end. Find the charge at the boundary of the conductors if a current I flows from conductor 1 to conductor 2. Sol. By Ohm’s Law, ⃗ = 

je

The problem can be solve in following ways: Find electric fields in both conductors.

iit

The current density in first conductor is

Similarly,

(Along the length of conductor)

J2 =

ac k

J1 =

along the length of conductor.

cr

The electric field inside first is E1 = 1 J =

Similarly,

E2 =

(Along length of conductor)

Now consider small cylindrical region at boundary The area of cylindrical region is S =  R2 The net flux through this region is  = 1 + 2 = ⃗⃗⃗⃗ . ⃗ + ⃗⃗⃗⃗ . ⃗ = E1 S cos 180 + E2 S cos 0 = - E1 S + E2 S = (E2 – E1) S

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According to Gauss’s Law, = or

(E2 – E1) =

=

∴ The surface charge density at boundary is  = (E2 – E1) Substituting the value E1 and E2, we get ∴

 = (2 j - 1j) or

∴

=

or

R2 =

or

r=

 = (2 - 1) j

(2 - 1)

e. in

(2 - 1) I (2 - 1) I

(2 - 1) I

q=

je

Charge at the boundary

ac k

iit

Q.10. The gap between the plates of a parallel plate capacitor is filled up with two dielectric layers 1 and 2 with thickness d1 and d2, permitivities 1 and 2, and 1 and 2. A dc voltage V is applied to the capacitor, with electric field directed from layer 1 to 2. Find , the surface density of extraneous charges at the boundary between the dielectric layers and the condition under which  = 0

cr

Sol. When electric field is uniform, then E=

This problem can be solved in following ways: Find electric field in both dielectric separately V = V1 + V2 or

(∵ E1 =

and E2 =

)

V = E1 d1 + E2 d2 …………………… (i)

If leakage current through capacitor is I, then J1 =

and

J2 =

According to Ohm’s law,

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E1 = 1J1

and

J1 =

J2 =



and

E2 = 2J2 

J1 = J2 =

∴



=

E2 =

  

E1

From eqn (i), we have

E1 d1 + E2 d2 = V 

E1 d1 + ∴

E1 =

Similarly,

E2 =



E1 d2 = V

 

 





je

Now apply Gauss’s law at the boundary:

e. in

But

iit

We consider a cylindrical region of cross sectional area dS near the boundary.

or

- 1 E1 dS + 2 E2 dS =

or

(2 E2 - 1 E1) =

or

cr

ac k

According to Gauss’s law, ∫ ⃗ . d ⃗ =

∵

(2 E2 - 1 E1) =



 = 0 (2 E2 - 1 E1)

Putting the values of E1 and E2, we get = For





(2 2 - 1 1)

 = 0, 2 2 = 1 1

Q.11. An in-homogenous poorly conducting medium fills up the space between the plates 1 and 2 of a parallel plate capacitor. It’s permittivity and resistivity vary

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from values 1, 1 at plate 1 to plate 2. Find the total extraneous charge in the given medium. Here, ρ varies linearly from first plate second plate. ∴

ρ = mx + K

When x = 0, ρ = ρ1 ∴

ρ1 = m × 0 + K

∴

K = ρ1

∴

ρ2 = md + ρ1 m=.

∴ є=.

Similarly,

ee .in

When x = d, ρ = ρ2 /x+є

/ x + є1

E = 2.

itj

The electric field at left surface of considered element is E = ρJ 3J

∵

dE = J d ρ

∴ ∴ Similarly,

cr ac

∵

ki

∴

/ E = ρJ

ρ=(

є=.

)x+

=.

/

=.

/

/

According to the solution of previous problem, σ= For considered element,

=

+d

= +d = ∴

dσ =

(

)(

)-

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+

= But

may be neglected.

∴

dσ =

or

dσ =

2.

2.

3.

/

3J.

/

= 2.

/

/ /

= .

/ 2.

/

-

-

3

/ 3+.

-

/

/

3.

3.

+

-

+

)

3 ) +.

/( -

3

/

3

/

/ 2(

+.

/

/

/ ∫ 2.

/ 2.

/

-

/

+

-

+

∴ Q=

-

3.

+.

) +.

+

/

/

3+.

cr

=

2.

)

+

-

/

+2.

/

/(

=.

=

.

/ 2(

=.

=

/

/ ∫ 2.

=.

∫

3.

2.

e. in

2.

Sdσ = I

or

or

/

je

or

3J .

J =

=. or

/

iit

or

+

ac k

But ∴

=

+

-

+

(

)

material whose Q.12 A long round conductor of cross-sectional area S is made of  resistivity depends only on a distance r from the axis of the conductor as ρ = , where  is a constant, Find: (a) The resistance per unit length of such a conductor. (b) The electric field strength in the conductor due to which a current I flows through it. ∵

=∫

Sol. Let us consider a long co-axial hollow cylinder of radius r and thickness dr

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The cross-sectional area of considered element is dA= 2

.

∴ Electric resistance between ends of conductor of considered element is dR = ρ

=ρ

=



The system may be assumed as parallel combination of a number of such types of elementary resistance.

∴

=

R=

∫

=

or

R=

. /

Resistance per unit length is

According to Ohm’s law

V = IR E=

=



=

ac k

But

. /

je

∴





S=

But

∴



e. in

=∫

iit

∴

E=. /I=



cr

∴

Q.1- A capacitor filled with dielectric of permittivity = 2.1 loses half the charge acquired during a time interval  = 3.0 min. Assuming the charge to leak only through the dielectric filler, calculate its resistivity. Sol. This system behaves as parallel combination of a capacitor C =

and a

resistance of R =

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∵ At t = 0, charge on capacitor is q0 When t > 0, charge start to leak because to this, charge on plate start to decrease. Applying loop rule in circuit. - IR = 0 I=

But

(since, charge decrease with

Or

= 0 or

+ ∫

∫

=

⁄

q = q0

ki

After integrating,

=

itj

∴

ee .in

increase of time)

Or ∴

= q0

⁄

cr

∴

ac

According to problem, when t = , q =

or

=

⁄

- ln 2 = =

After substituting the values, we get



= 1.4× 1013  m

Q. 2 a circuit consist of a source of a constant emf  and a resistance R and a capacitor with capacitance C connected in series. The internal resistance of the source is negligible. At a moment t = 0 the capacitance of the capacitor is

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abruptly decreased -fold. Find the current flowing through the circuit as a function of time t. Sol. This problem is solved by following ways: First draw circuit at t < 0.

ee .in

In this situation, circuit is in steady state. So, current in the circuit is zero. According to the loop rule,

- =0 Q=C

itj

∵

i.e.

cr ac

abruptly decreased.

ki

Now discuss the situation at t = 0. Since, at t = 0, the capacity of capacitor is

C1 =



Due to this, voltage of the capacitor abruptly increases. This voltage is greater than the emf of cell. So, charge starts to decrease for getting new steady. The direction of flow of charge

Now draw the circuit at an instant t (> 0). At an instant t, the circuit According to loop rule,

– IR -  = 0

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Here

I=

(since charge decreases)

∴

-

Or

C1 – q = RC1

Or

∫

Or

[

=R



–



–



–



–

=

je

–

1=

= (

–

)

⁄

iit

Or

 - q = .

Or



ac k

=

e. in



ln

–

– )]Q =

0

Or



∫

=

(

Or

Or

–

or



 - q= . 

⁄

– / 



⁄

/

cr

Differentiating both sides with respect to t = ( ) . 

I = (

Or

)

 



/. /

⁄

⁄

Q.3- Find the potential difference 1- 2 between two points 1 and 2 of the circuit. If R1 = 10, R2 = 20, 1 = 5.0 V and 2 = 2.0 V. The internal resistances of the current sources are negligible.

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Sol. ∵ The potential difference between two points = 1- 2 = - {algebraic sum of rise up and drop up of the voltage} Here, the distribution of the current

According to loops rule,(in loop ABDEA)

∴

I=





e. in

-IR1 + 1 – IR2 + 2 = 0

je

∴ 1- 2 = - {algebraic sum of rise up and drop up of the voltage throw path (1A

iit

B2)}.

ac k

=-{-IR1 + 1} = IR1 - 1

Substituting the value of I, we get

(

 )



cr

1- 2 =

Substituting the values, we get, 1- 2 = -4V Q.4- Two sources of current of the equal emf are connected in series and have different internal resistance R1 and R2, (R1 > R2). Find the external resistance R at which at which the potential difference across the terminals of one of the sources (which one in particular?) becomes equal to zero. Sol. Here the circuit of the problem.

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According to the loop rule,  - IR1 +  - IR2 – IR =0 ∴



I=

Potential difference between the terminals of first cell is A- B = - {

}

A- B =

Or



e. in

According the problem,



=

A- B = 0 

iit =1

2 R1 =

or R =

R2 > R1

cr

∴

ac k

Or Or

=0

je

∴

So R is negative, which is physically not possible. The potential difference across the second cell is B- C = - { Or

B- C = =

} 





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Q.5- N source of current with different emf are connected. The emf of the sources are proportional to their internal resistances, i.e.  = R, where  is an assigned constant. The lead wire resistance is negligible. Find

(a)

The current in the circuit;

(b)

The potential difference between the points A and B dividing the circuit

e. in

in n and N –n links.

Sol. By loop rule, (

) + (

) + ……….+upto N terms = 0

je

(a)

iit

N - NIR = 0

∴

A- B = ( 

cr

(b)







= = =

ac k

I=

)+( 

) + ………+ upto n terms

= n - nIR = nR - nR = 0

Q.6- In the circuit, the source have the emf’s 1 = 1.0V and 2 = 2.5V and the resistance have the values R1=10 and R2 = 20. The internal resistances of the source are negligible. Find a potential difference A- B between the plates A and B of the capacitor C.

Sol.

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Since, in steady state, the current through the capacitor branch is zero. Also charge on capacitor is maximum. The current distribution in the circuit. Form KVL, in closed loop ABEFA, 1 – IR1 –IR2 -2 =0 ∴

I=





G- D = {1 – IR2}

e. in

= 2

3

1 + IR1 = 2

∴

= -IR1=.

(through capacitor branch)

je

Or





iit ac k

∴

(through GABD)

A- B = 2

/ R1

3 =.





/ R1

cr

=

3

By substituting the values, we get A- B = -0.5 V Q.7- In the circuit the emf of the source is equal to = 5.0V and the resistance are equal to R1 = 4.0 and R2 = 20. The internal resistance of the sources equals R = 0.10. Find the current flowing through the resistance R1 and R2. Sol. The distribution of current in the circuit:

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By KVL, in closed loop AGEFA,  – IR –I1R1 = 0 –

∴

I=

In loop GBDEG,

-(I- I1) R2 + I1R1 = 0 ……………………(ii)

…………………………(i)

After substituting the value of I, we get 

From equation (ii), we get



iit

Substituting the value of I1, we get

je

I –I1 =

= 1.2A

e. in

I1 =

ac k

I –I1 =

On substituting the value we get,

cr

I2 = 0.8A

Q.8- Find the current flowing through the resistance R in the circuit. The internal resistances of the batteries are negligible. Sol. Here the distribution of the current in the circuit is In loop (1), 0 – (I –I1) R1 = 0

………(i)

In loop (2), -I2 R3 –I1R2 + (I –I1) R1 = 0

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……..(ii)

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In loop (3),  + I2 R3 – (I1 -I2) R = 0

………(iii)

By solving equation (i), (ii), (iii), we get I1 –I2 =

) 

( (

……..(iv)

)

Q.9- Find a potential difference A- B between the plates of a capacitor C in the circuit shown in . If the sources have the emf’s 1 = 4.0V and 2 = 1.0V and the

of the sources are negligible.

e. in

resistance are equal to R1= 10, R2 = 20, and R3=30. The internal resistances

Sol. The current through the capacitor branch is zero in steady state. The

iit

In loop (1), (DEFGKD)

je

distribution of the current in the circuit.

cr

In loop (2)(GHJKG)

ac k

1 –IR1 –I2R2 = 0

2 – (I –I1) R3 + I1R2 = 0

…….(i)

……..(ii)

After solving equation (i) and (ii), we get I=

 (

) 

Potential difference between the points F and H through paths FGH. F- H = - {algebraic sum of rise up and down up of voltage} = - {-IR1+2} = IR1 -2 By substituting the values of I, we get

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F- H = H- F =

Or



(

) 

(

)



(

) 

(

)

Form F = B, H = A ∴

B - A = H- F =



) 

(

(

)

By substituting the values, we get

ee .in

B - A = -1V Q.10- Find the magnitude and direction of the current flowing through the resistance R in the circuit shown in . If the emf’s of the sources are equal to  =

itj

1.5V and  = 3.7 V and the resistance are equal to R1 = 10, R2 = 20 and R =

ki

5.0. The internal resistances of the sources are negligible.

cr ac

Sol. The distribution of current in the circuit In loop (1), from KVL

 



Or





∴

I=

Or



(





)

=0 =0

(

)

+.

/

=0 ……(i)

In loop (2), ( Or

(

) )

 =0  =0

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∴

=



Substituting the values of I from (i) and then solving, we get, 



From point A to B On substituting the values, we get = 0.02A

e. in

From A to B i.e. from left to right.

Q.11- In the circuit shown in the the sources have the emf’s  = 1.5V and  =

je

2.0 V and  = 2.5V, and the resistance are equal to the R1 = 10, R2 = 20 and R3

iit

= 30. The internal resistances of the sources are negligible. Find;

ac k

(a) The current flowing through the resistance R1;

cr

(b) A potential difference, A- B between the points A and B.

Sol. We know that potential difference between two points is A - B = {algebraic sum of rise up and down up of voltage between points A and B through any path} The distribution of current in the circuit From KVL, in loop (1) 

(

)

 =0

……(i)

In loop (2),

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 (a)

(

)

=0

…….(ii)

After solving the equation (i) and (ii), we get =

(b)



(

 )

(

 )

= 0.6A

………(iii)

Taking the potential difference between point A and B through the path AEFB,

= Putting the values of

}

e. in

A - B = { 

from equation (iii), we get,

je

A - B = 0.9V

iit

Q.12- A constant voltage V = 25V is maintained between the points A and B of the

ac k

circuit. Find the magnitude and direction of the current flowing through the segment CD if the resistances are equal to R1 = 1.0, R2 = 2.0 and R3 = 4.0.

cr

Sol. For the sake of simplicity an ideal battery of emf V= 25 volt is connected between the point A and B. From KVL In loop HJADGBH, V - (I- I1) R3 – (I -I1 -I3) R4 = 0

……..(i)

In loop ECDAE, -I1R1 + (I –I1) R3 =0

……….(ii)

In loop CFBGDC,

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(I1 –I3) R2 + (I –I1 –I3) R2 = 0 ………(iii) After solving the equation (i) 2

I3 =

,

)⁄

(

(

)-

3

After substituting the values, we get I3 =1A

e. in

Q.12- Find the resistance between the points A and B of the circuit shown in . Sol. This concept is based upon rotator symmetry. If a circuit is rotated through 1800 about a point O and no change takes place in the dimensions of circuit takes

iit

je

place in the dimension of circuit. Then circuit is in rotator symmetry at that point.

ac k

Here, if the circuit shown in the is rotated through 1800 about the middle point O of the line CD, then no change takes place in the dimensions of circuit.

cr

So, the circuit is in rotational symmetry about point O. Due to rotation, branch CE is same as that of in FD, distribution of current is: In loop (1), (JKAFDGBJ) V – (I –I1) R – I1r = 0

…..(i)

In loop (2),(ECDAFE) -I1r – (2I1 – I) r + (I –I1) R = 0

……(ii)

After solving the equation (i) and (ii), we get

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I=

(

)

(

)

……..(iii) Now the circuit between the point A and B may be replaced by an equivalent resistance RAB. The equivalent circuit According to KVL, V – IRAB = 0

From equation (iii) and (iv), we get ( (

)

)

je

RAB =

e. in

……..(iv)

iit

Q. 13 – Find how the voltage across the capacitor C varies with time t after the

ac k

shorting of the switch Sw at the moment t = 0.

cr

Sol. The question is an example of R – C circuit in parallel. Here the distribution of the current and the charge in the circuit at an instant t From KVL, in loop (1), (ABFGA),  - (I –I1) R –IR = 0 Or

 - IR + I1R – IR = 0

Or

 - 2IR + I1R = 0

∴

I=



……..(i)

In loop BDEFB,

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)

( or

.



or

.



or

.





or

=0 /

=0

/

=0

/=0

=

ee .in

But ∴



or

C - 2q =

itj

=

cr ac

After integrating we get,

q=

or ∴

=∫



ki

∫

or



(

⁄

)



(

⁄

)



(

⁄

)

=

V=

Q.14- What amount of heat will be generated in a coil of resistance R due to a charge g passing through it if the current in the coil (a) decreases down to zero uniformly during at time interval t; (b) decreases down to zero halving its value every t seconds ?

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Sol. ∵ Electric current I= q=∫

Also,

and, heat is generated in circuit is ∫

Here, current decreases uniformly with time. It means current is linear

e. in

(a)

for variable current

function of time. ∴

je

I = mt + C

∴ ∴

ck

∴

I0 = m × 0 + C

cr a

At t = 0, I = I0

iit

When m and C are constant

C = I0 I = mt + I0

At t = , I = 0 ∴

0=m

∴

m=

∴

I=

∴

I = I0

+ I0

+

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But total charge transferred through circuit in the time ∴

=∫ .

q=∫

/

=0

∴

is q.

10 =

=

I0 =

∴ But =∫ .

=∫ .

/

After integrating we get, =

je

Let at t = 0, current is I0

iit

(b)

=

/

e. in

=∫

∴

ac k

At t = , I =

I=

. /

=

cr

At t = 2

Similarly, at t =

I= ∴

I=

. / . / =

. /

⁄

(∵ t=

In this case, current becomes zero after infinitely large time. q=∫

=∫

. /

⁄

dt

After integrating, we get

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q= ∴

=

∴

=

=∫

= ∫

.

=

=

=∫ / . /

. /

⁄

Rdt

⁄

e. in

After integrating, we get

Q.14- A dc source with internal resistances R0 is loaded with three identical

je

resistances R interconnected as shown in . At what value of R will the thermal

ac k

iit

power generated in this circuit be the highest?

From KVL or loop rule in the circuit ABEFA,

∴

cr

 -IR0



I=

=

=0 

The power in the circuit is P=

. /=.



/ . /=(

 )

For maximum power generated in the circuit, is =0

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After solving, we get ∴ R = 3R0

R0 =

Q.15- Make sure that the current distribution over two resistances R1 and R2 connected in parallel corresponds to the minimum thermal generated in this circuit.

∵ The value of resistance is always positive. In parallel combination of two

e. in

resistance, i  =

je

i.e.

iit

The thermal power dissipated in resistance is P =

ac k

Total power in the circuit is, P = or

or

cr

or

or or or or

P=

(

P=

(

P= (

)[ P=(

)

+

P= P=(

)

) .

/

)0 P=(

.

/

]

1 )0

1

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.

/

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for being P minimum,

P=0



0

∴

1 should be minimum 1 =0

=

for minimum power dissipated.

Q.16- A storage battery with emf  = 2.6 V loaded with an external resistance produces a current I = 1.0 A. In this case the potential difference between the

e. in

terminals of the storage battery equals V = 2.0 V. Find the thermal power generated in the battery and the power develops in it by electric forces. Sol.

je

From KVL

iit

 -I r – IR = 0

∴ ∴

…….(i)

A - B = V = - {- + I r}

cr

Also,



ac k I=

V= -Ir r=



………(ii)

The thermal power generated in case is due to internal resistance. ∴

P=

∴

P = I (

=

.



/

)

On substituting the values, we get

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P = 0.6 W The net power developed by electrical forces is P1 = -V1 = -2 W Q.17-A voltage V is applied to a dc electric motor. The armature winding resistance is equal to R. At what value of current flowing through the winding will the useful power of the motor be the highest? What is it equal to? What is the

e. in

motor efficiency in this case?

Sol. In the circuit, here motor can be assumed as a resistance.

je

Here V = D.C. applied voltage

∴

cr

From KVL,

ac k

R0 = Resistance of motor

iit

R = resistance of the winding armature

V – IR – IR0 =0 I=

∴The power dissipated in motor is P=

=(

)

For maximum power dissipated, =0 After solving, we get

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∴

I=

∴

P max =

=

(

=( ∴

)

)

P max =

R=

(∵ R = R0)

e. in

The power supplied by D.C. voltage is

Percentage efficiency is

ac k

iit

=

/=

je

Pi = VI = .

=4 5

=

(∵ R = R0)

=

= 50%

cr

Q.18- How much (in percent) has a filament diameter decreased due to evaporation if the maintenance of the previous temperature required an increase of the voltage by  = 1.0%? The amount of the heat transferred from the filament into surrounding space is assumed to be proportional to the filament surface area. Sol. At steady state, internal energy of filament becomes constant. So, the temperature becomes constant. From energy point of view, thermal power developed in filament is lost at heat in environment at steady state.

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According to given problem, thermal power Q loss is proportional to the surface area A of filament (cylindrical in shape) mathematically, Q A

or Q = kA or Q = k 2rl

Where r = radius of filament and l = length of filament The power supplied by the applied voltage V is P=

∴

ee .in

At steady state, P=Q

= k 2rl

or

or

cr ac

From equ. (i), we get ∴

=

= k 2rl = k 2rl

or

= 2k

or

= k0

∴

=

ki

itj

Here R = resistance of the filament

………..(i)

(here k0 = 2k

= constant)

V2 =

Differentiating both sides with respect to r, 2V

=-

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Negative sign indicates that when radius of filament decreases, voltage increases. ∴

2V .

or

2V

=

or

2V

=

or

2

=

or

2

=

/ = .∵ .∵

/

ee .in

=

100 = 2 .

∴

/

/

Hence, percentage decrease in diameter of filament is

itj

/ = 2η

ki

100 = 2.

ac

= 2 × 1% = 2%

Q.1. A conductor has a temperature –independent resistance R and a total

cr

capacity C. at the moment t = 0 it is connected to a dc voltage V. Find the time dependence of a conductor’s temperature T assuming the thermal power dissipated into surrounding spaces to vary as q = k(T - T0), where k is a constant, T0 is the environmental temperature (equal to the conductor’s temperature at the initial moment). Sol. It is understood that before steady state, power supplied = Thermal power loss + change in internal energy.

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When a conductor receives electrical energy internal energy of conductor increases so, the temperature of conductor increases at t = 0, no temperature differences are found between conductor and environment. So at t = 0, heat loss is zero. When t > 0, a part of electrical energy increases internal energy (temperature of conductor) and remaining part of electrical energy is lost as heat. Due to this, temperature difference between conductor and environment

e. in

increases. Hence, heat lost increases. After some time, steady states come into play. At steady state, total electrical

je

power supplied to conductor is lost as heat in environment.

iit

The temperature of conductor becomes constant.

ac k

Before steady state:

energy

cr

Electrical power supplied = Thermal power lost + rate of change in internal

or

= k (T - T0) +

or

or

= k (T - T0) + C

or

= k (T - T0) C∫

(

(CT)

)

=∫

After integrating, we get T=

(1-e-kt/c)

Q.2. A circuit shown in has resistances R1 = 20 and R2 = 30 . At what value of the resistance Rx will the thermal power generated in it be practically

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independent of small variations of the resistance ? The voltage between the points A and B is supposed to be constant in this case. Sol. Power developed in resistance Rx to be maximum, the power generated in Rx is practically independent of small variation of resistance. For maximum power developed in Rx is, =0

V - IR1 - (I - I1) R2 = 0

e. in

In loop (1), ……(i)

- I1RX + (I - I1) R2 = 0

ck

iit

After solving Eq. (i) and (ii), we get

…….(ii)

je

In loop (2),

I1 =

cr a

The thermal power generated in the resistance Rx is P = I12 RX = .

For maximum value of p,

/2 RX

=0

After solving, we get RX = On putting the values, we get RX = 12 Q.3. in a circuit shown in . resistances R1and R2 are known, as well as emf’s 2.

1 and

The internal resistances of the sources are negligible. At what value of the

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resistance R will the thermal power generated in it be the highest ? What is it equal to? Sol. This problem can be solved by maximum power transfer theorem. The problem can be solved in following ways: First: Show the original circuit. Now : remove the resistor of resistances R from the circuit In this step, we remove that resistance from the circuit in which maximum

e. in

thermal power is generated.

Now: drew the circuit after short circuited the all cells(but internal resistances of

je

cells remains in the circuit) of circuit.

iit

The circuit of this step

cr

RBD =

ac k

The equivalent resistance of circuit in between point B and D is

For maximum power generated in resistance R,the value of R. = RBD = Now draw original circuit and find current through R. In loop (1),

1-

IR1 - I1R = 0

….(i)

In loop (2) - (I - I1)

2 + I 1R

=0

….(ii)

After solving Eq.(i) and (ii), we get

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I1 =

)

(

∴ Power generated in resistance R is P = For Pmax, ∴

R

R = RBD = Pmax =

) (

)

(

.

/

) (

)

e. in

=

(

Q.4. A series-parallel combination battery consisting of a large number N = 300

je

of identical cells, each with an internal resistance r = 10 Ω. Find the number n of parallel groups consisting of an equal number of cells is connected in series, at

iit

which the external resistance generates the highest thermal power.

ac k

Sol. This problem can be solved in the following way First: find equivalent emf and internal resistance of the combination of cells. Let

cr

each row contains number of shell

Let each row contains number of cells in series. ∴

The equivalent emf of each row is

0=

n0

Also, the equivalent internal resistance of each row is r0 = n0r The equivalent circuit is Now connect the equivalent cell with external resistance R.

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From KVL 0-

– IR = 0

∴ or ∴

I= I=

(∵ =

and

0=

)

I=

∴

n0 = . /

=

iit

I=

je

∴

N = nn0

e. in

The total number of cells = N = number of row × number of cells in each row

ac k

The power generated in external resistance R is P=

/ R

=0

cr

For maximum power,

R=.

After solving, we get

n =√

On putting the values, we get n = 3 Q.5. A capacitor of capacitance C = 5.00 F is connected to a source of constant emf  = 200 V. Then the switch Sw was thrown over from contact 1 to contact 2. Find the amount of heat generated in a resistance R1 = 500  if R2 = 330 . Sol.

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This problem can be solved in following ways : Now draw the circuit when switch is at position at 1. The circuit is shown in . In this case, the circuit is in steady state. So, current in the circuit will be zero. In loop (1) =

∴

=0

q0 = C  ………..(i)

e. in

Now discuss the circuit at the position (2) of the circuit : At t = 0, when the switch Sw was thrown over from contact 1 to contact 2, current in the circuit is zero. Now capacitor starts to discharge through resistances R1 and

je

R2.

iit

After long time, again circuit comes in steady state. At this steady state, charge

ac k

on capacitor and current in circuit both are zero. From energy point of view energy stored on capacitor appears as thermal energy

cr

in resistors R1 and R2.

But R1 and R2 are in series, so current in both are same. ∴

Δ H = I2 Rt

∴

=

∴

∴ Δ H1  R1 and

Δ H2 =

Δ H2  R2

Δ H1 ……………………….. (ii)

Total heat generated is

Δ H = Δ H 1 + Δ H2

According to conservation principle of energy, Δ H + Δ U = 0

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or

Δ H + (Uf - Ui) = 0

∴

Δ H = Ui – Uf

∴

ΔH=

or

Δ H1 + or ∴

Δ H1 .

Δ H1 + Δ H 2 = Δ H1 =

/= Δ H1 = (

(from eqn (ii)) Δ H1 .

or

)(

)

=(

/=

)

( )(

)

=

(

)

= 60 mJ

e. in

-0

Q.6. A glass plate totally fills up the gap between the electrodes of a parallel plate

je

capacitor whose capacitance in the absence of that glass plate is equal to C = 20 F. The capacitor is connected to a dc voltage source V = 100 V. The plate is

iit

slowly and without action, extracted from the gap. Find the capacitor energy

ck

increment and the mechanical work performed in the process of plate extraction.

cr a

Sol. From conservation principle of energy, we can write Wext + Wcell = Δ U

Here, Wext = The work done by extern agent (mechanical work) against electrical forces. Wcell = work done by cell

= V Δq

Δ U = Ui – U f The capacity of capacitor without glass plate = C = The capacity of capacitor with glass plate is C1 =

= C

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The initial energy stored on capacitor is U i = C1 V 2 =  C V 2 The charge on capacitor is qi = C1V =  CV The energy stored on capacitor after removal of glass plate is Uf = CV2

qf = CV

e. in

The charge on capacitor after removal of plate is

Δ U = Ui – Uf = CV2 (1 - )

Also,

Δq = qf – qi = CV - CV = CV (1 - ) < 0

∴

Wcell = V Δq = CV2 (1 - ) < 0

iit

je

∴

ac k

It means work done by cell is negative.

cr

From conservation of energy, Wext + Wcell = Δ U Wext = Δ U - Wcell = CV2 (1 - ) - CV2 (1 - ) = - CV2 (1 - ) = CV2 ( - 1) > 0 For glass

=6

On putting the values, we get Wext = 0.5 mJ The increment in energy of capacitor is

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Δ U = Ui – Uf = - CV2 (1 - ) = - 0.5 mJ Q.7. A cylindrical capacitor connected to a dc voltage source V touches the surface of water with its end. The separation d between the capacitor electrodes is substantially less than their mean radius. Find a height h to which the water level in the gap will rise. The capillary effects are to be neglected. Sol. ∵ The capacity of capacitor without water is C0 =  0 .

ee .in

/

Let a liquid rises up to height h in the capacitor.

C=



) 

+

 

[l + ( - 1) h]

cr ac

∴

(

ki

=

itj

C = Cair + Cliquid

where  is dielectric constant of liquid. ∵ The initial energy of capacitor is U i = C0V 2 And the final energy of capacitor is Uf = CV2 ∴ Increase in energy of capacitor is Δ U = Uf – Ui = - V2 (C – C0)

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Work done by cell = V Δq = V (qf - qi) = V (CV – C0V) = V2 (C – C0) Increase in gravitational potential energy = Δ Ug = mg =  × volume × .g = g (2Rhd) × = gRh2d

ee .in

According to conservation principle of energy, a part of work done by cell increases potential energy of capacitor and remaining part of work done by cell

Wcell = Δ U + Δ Ug

Putting the values, we get (

)

ac

h=

ki

∴

itj

increases gravitational potential energy of liquid.

cr

Q.8. The radii of spherical capacitor electrodes are equal to a and b, with a < b. The inter-electrode space is filled with homogenous substance of permittivity  and resistivity . Initially the capacitor is not charged. At the moment t = 0, the internal electrode get a charge q0. Find : (a) the time variation of the charge on the internal electrode; (b) the amount of heat generated during the spreading of the charge. Sol. This system behaves as R – C discharging circuit.

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Here the system may be assumed as parallel combination of a resistance and capacitance between capacitor electrodes. The problem can be solved by following ways: First find equivalent resistance and capacitance between electrodes. As we know electric resistance between electrode is R=

.

(

/=

)

e. in

Also, the capacity of spherical capacitor is C=

je

Now draw equivalent circuit :

iit

At t = 0, charge on capacitor is q0. Now capacitor starts to discharge through

ck

resistance R.

an instant t.

cr a

Let at an instant t, charge on capacitor is q and current in circuit is I. The circuit at

Since, charge on capacitor decreases with respect to time ∴

I=

=

According to loop rule, - IR = 0 +R

or or

∫

=0 =∫

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After integrating, we get q = q0 e-t/RC Substituting the values of R and C, we get q = q0 e-t/0 q = q0 e-t/RC I=∴

e-t/RC

= 

ΔH=∫ =∫

 (

Rdt

e-t/RC Rdt

)

)

(

ki

=

itj

After integrating, we get ΔH=

ee .in

(b)

cr ac

Q.9. The electrodes of a capacitor of capacitance C = 2.00 F carry opposite charges q0 = 1.00 mC. Then the electrodes are interconnected through a resistance R = 5.0 m. Find:

(a) the charge flowing through that resistance during a time interval = 2.00 sec; (b) the amount of heat generated in the resistance during the same interval. Sol. The circuit of problem is Let at an instant, charge on capacitor is q and current in circuit is I. From KVL,

- IR = 0

or

-R.

/=0

.∵

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or

- RC ∫

∴

=∫ q = q0 e-t/RC

When t =

q = q0 e-/RC

Here q is charge on capacitor at t = . The charge flow through circuit is equal to the decrease in charge on capacitor. ∴ The charge flow through circuit is Q = q0 – q = q0 (1- e/RC)

e. in

On putting the values, we get Q = 0.18 mC (b) ∵

e-t/RC

je

=

iit

∴

I=-

= ∫ . / e-2t/RC Rdt

H=∫

ac k

or

q = q0e-t/RC

H=

cr

After integrating, we get

On substituting the values, we get

(1 - e-2/RC) H = 82 m

Q.10. In a circuit shown in the capacitance of each capacitor is equal to C and the resistance, 15 R. One of the capacitors was connected to a voltage V 0 and then at the moment t = 0 was short circuited by means of the switch SW. Find: (a) a current I in the circuit as a function of time t ; (b) the amount of generated heat provided a dependence I (t) is known. Sol. This problem can be solved in following ways:

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First: Show the position of circuit before short circuited by means of the switch SW. the circuit is In this case, the capacitor is in steady state. Applying KVL

V0 -

=0

∴

q0 = CV0

………….(i)

Now: show the position of circuit at an instant t:

that of second capacitor is increase. - IR - = 0 -

or

(q0 - 2q) = RC

cr ac

or or

∫

=∫

ln

=-

or

= e-2t/RC

∴

q=

(1 - e-2t/RC)

e-2t/RC =

e-2t/RC

(a) ∴ (b)

/

ki

or

.∵

=0

–R

itj

From KVL,

ee .in

When switch SW is closed at t = 0, change on first capacitor starts to decrease. But

I=

=

H=∫ =

Rdt ∫ -4t/RC dt =

(∵ q0 = CV0) [- e-4t/RC + 1]

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∴

[1 - e-4t/RC]

H=

Since, current is flowing for long time ∴

∴t

H=

(∵ q0 = CV0)

= CV0

Q.11. A coil of radius r = 25 cm wound of thin copper wire of length l = 500m rotates with an angular velocity

= 300 rad/s about its axis. The coil is connected

to a ballistic galvanometer by means of sliding contacts. The total resistance of

ee .in

the circuit is equal to h = 21 . Find the specific charge of current carries in copper if a sudden stoppage of the coil makes a charge q = 10 nC flow through the galvanometer.

itj

Sol. From Faraday’s law of electromagnetic induction



cr ac

IR = - N

ki



=-N

R

∴

=-N R

∴



=-N

q=N

Here

|

|

= net charge in magnetic flux, r = total resistance of coil.

q = total charge transferred. Initial flux is i = r2B per turn. Final flux is ∴

F = 0 = f - i = - r2B

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∴

q=N

|

|

=

……..(i)

Since, electron in coil moves on circular path with constant angular velocity For moving on circular path, magnetic force on electron provides required centripetal force. evB =

∴

B=

or =

eB =

=

e. in

∴

Here m = mass of electron, e = charge on electron. ⃗ =-

∴

Δq=

(∵ Charge on electron is negative)

ac k

iit

⃗⃗

je

In vector form,

On substituting the value of B, we get .

/

cr

= =

∴

( ∵ The average flow of charge is = )

=

The number of turns

N=



=. =



/

= 1.8 × 1011 C/kg

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Q.12. Find the total momentum of electrons in a straight wire of length l = 1000 m carrying a current I = 70 A. ∵ The drift velocity of electron is vd = where J = current density n = number of electrons per unit volume

e. in

e = charge on electron Sol. The momentum of each electron is P0 = m vd

je

Here m = mass of each electron = 9.1 × 10-31 kg

iit

Then the total momentum of electrons inside the conductor is

ac k

P = NP0

Let S = cross-sectional area of straight wire (cylindrical wire) J=

cr

∴

Volume of wire is

V = Sl

∴ no. of electron inside the conductor is N = nSl ∴

P0 = NP0 = (nSl)

∴

P = (nSl)

=

= 0.40 Ns

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Q.1. A straight copper wire of length l = 1000 m and cross-sectional area S = 1.0 mm2 carries a current I = 4.5 A. Assume that one free electron corresponds to each copper atom. Determine (a) the line it takes an electron to displaced from one end of the wire to the other; (b) the sum of electric forces acting on all free electrons in the given wire. Sol. (a) The time taken by electron to displace from one end to another end is =

=

=

t = 3 ms

je

On substituting the values, we get

=

e. in

t=

iit

(b) F = qE

ac k

Here q = total charge on electron = Ne

Where n = number of electrons per unit volume. According to Ohm’s law in vector form,

cr

E = J

Where  is resistivity of copper. ∴

F = qE = NeE F = NeJ = nSleJ ∵ F = nSleI

F = nSle On substituting the values, we get F = 1 MN

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Q.2. A homogenous proton beam accelerated by a potential difference V = 600 kV has a round cross-section of radius r = 5.0 mm. Find the electric field strength on the surface of the beam and the potential difference between the surface and the axis of the beam if the beam current is equal to I = 50 mA. Sol. For analyzing the problem, the homogenous proton beam may be assumed as a long non-conducting cylinder of radius r.

ee .in

Now, we consider a point P at distance x (x < r) from the axis of cylinder. Let the volume charge density of the beam is . For calculating electric field at point P, we draw an imaginary Gaussian co-axial cylinder of length l passing through point

itj

P.

E 2xl =



∴E=

cr

or



ac

E (2xl) =

ki

According to Gauss’s law, ∫ ⃗ . d ⃗ =

Calculating for  :

Let the speed of each proton is u. ∴

eV = mu2

where m is mass of proton. e = charge on proton

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∴

u=√

∴ I=

Here

dq =  dV =  r2 dx

∴

I=

= r2

= r2u

= 

Electric field at the surface is, E =

=



x



∴

E=

∴

E=

Here

u=√

∵

.∵





/

ki

itj



E= E=

√



= 32 V/m



-∫

=∫ 

- ∫ ∴



cr ac

∴

=

ee .in

For surface, x = r

=∫

1 - 2 =



dx



On substituting value of , we get Δ  = 1 - 2 =



√

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Q.3. Two large parallel plates are located in vacuum. One of them series as cathode, a source of electrons whose initial velocity is negligible. An electron flow directed toward the opposite plate producing a space charge causing the potential in the gap between the plate to vary as =ax3/4, where a is positive constant, and x is the distance from the cathode. Determine; (a) the volume density of the space charge as a function of x; (b) the current density.

+

=

 

 = a x4/3

je

Here

+

e. in

∵ We know

E=-

and

+

E = - x1/3

+



=



cr

∵ we know,



ac k

∴

iit

sin  = f(x)

Here E is only the function of x. ∴

= 0,

=0 ∵

∴ or or

/=

. - a

=

 

 

=

 

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=-

∴



x-2/3

(b) Now, we consider an element of thickness dx at distance x from cathode. The electric charge in considered element is dq =  Sdx ∴

I=

= S

= Su

eV = mu2

As we know,

e. in

Electric current density is J = = u

Where m is mass of electron and V is potential difference between point P and

je

cathode.

∴

u=√

∴

cr

ac k

 = a x4/3

iit

∴ Potential difference between point P and cathode is

J = u = 0 a3/2 √

Q.4. A gas is ionized in the immediate vicinity of the surface of plane electrode 1 separated from electrode 2 by a distance l. An alternating voltage varying with time t as V = V0 sin t is applied to the electrodes. On decreasing the frequency , it was observed that the galvanometer G indicates a current only at ω < ω0, where ω0 is a certain cu-off frequency. Find the mobility of ions reaching electrode 2 under these conditions.

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Sol. ∵ vd = uE = u

(∵E= )

vd =

or

vd, 0  ω t  

For positive value of

∴ The collision of ion with electrode is due to current in the circuit. If maximum displacement of an ion is equal to separation l between plates, then

For this,

xmax = l at ω = ω0 vd = 0

∴





∵ vd =

or

= ∫

=

∫



cr

or

ω0 = t

iit

t=

=0

ac k

∴

vd =

je

∵ At maximum displacement,

e. in

deflection in galvanometer is observed.

After solving, we get l= ∴

u=

Q.5. The air between two closely located plates is uniformly ionized by ultraviolet radiation. The air between the plates is equal to V = 500 cm3, the observed saturation current is equal to Isat = 0.48 A. Determine

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(a) The number of ion pairs produced in a unit volume per unit time; (b) The equilibrium concentration of ion pairs if the recombination coefficient for air ion is equal to r = 1.67 × 10-6 cm3/s. At saturated current, all ions produced (positive ions + negative ions) reach at plate. Sol. ∵ The cause of current is motion of ions (

)

e. in

I= =

je

ni = =

iit

On putting the values we get

ac k

ni = 6 × 109 cm-3 s-1 (b) The recombination rate of ion is rn2

ni =

cr

The production rate of ion is ∴ Rate of increase of ion is =

= ni – r n2

For balance condition ∴

=0 n=√

On putting the values, we get n = 6 × 107 /cm3

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Q.6. Having been operated long enough, the ionizer producing n1 = 3.5 × 109 cm-3 s-1 of ion pairs per unit volume of air per unit time was switched off. Assuming that the only process tending to reduce the number of ions in air is their recombination with coefficient r = 1.67 × 10-6 cm3/s, find how soon after ionizer’s switching off the ion concentration decreases  = 2.0 times. Sol. From previous analysis, we know n0 = √

ee .in

at t = 0

The recombination rate is r n2

Since the ionizer producing is switched off ∴

itj

= - r n2 √

=-∫

ki

∫√



ac

or After solving,

t=

√

= 13 ms

cr

Q.7. A parallel plate air capacitor whose plates are separated by a distance d = 5.00 mm is first charged to a potential difference V = 90 V and then disconnected from a dc voltage source. Find the time interval during which the voltage across the capacitor decreases by  = 1.0%, taking into account that the average number of pairs formed in air under standard conditions per unit volume per unit time is equal to ni = 5.0 cm-3s-1 and that the given voltage corresponds to the saturation current.

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Sol. ∵ Because of production of ions, charge of capacitor starts to leak. This is caused by decrease of voltage of capacitor. C= And

I=

or

I=

(

)

)

.∵

I t = CV – CV’

or

t=

or

t=

(V – V’) ………………… (i) (

) )

(

)

)

=

ac k

According to problem,

(

=

je

(

e. in

Flow of charge in time t = decrease in charge on capacitor in time t.

or

∴

/

iit

∴

(

=

V – V’ = V

cr

∴ From equation (i), we have t=



on putting the values, we get t = 4.6 days Q.8. The gap between two plane plates of a capacitor equal to d is filled with a gas. One of the plates emits v0 electrons per second which is moving in a magnetic field, ionize gas molecules. In this way, each electron produces  new

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electrons (and ions) along a unit length of its path. Find the electric current at the opposite plate, neglecting the ionization of gas molecules formed by ions. Sol. ∵ Here number of new electrons in unit length is = v ∫

or

ln

=∫ =x = ex

∴ For saturation current,

x=d

itj

I = ve = ev0 ed

ee .in

or

ki

Q.9. The gas between the capacitor plates separated by a distance d is uniformly ionized by ultraviolet radiation so that ni electrons per unit volume per second

cr ac

are formed. These electrons moving in the electric field of the capacitor ionized molecules, each electron producing  new electrons (and ions) per unit length of its length of its path. Neglecting the ionization by ions, find the electronic current density at the plate possessing a higher potential. Sol. ∵ We know that v = v0 ex Here,

v0 = (ni S dx)

∴

v = (ni S dx) ex

∴

I=∫

=∫

dx

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On substituting the values, we get I = e ni S . ∴

/

(ed – 1)

J= =

Q.10. A current I = 1.00 A circulates in a round thin wire loop of radius R = 100 mm. Find the magnetic induction (a) at the centre of the loop

ee .in

(b) at the point lying on the axis of the loop at a distance x = 100 mm from its centre.

Sol. ∵ The magnetic field at the centre of a circular arc is   

ki

Where I = current in arc

itj

B=

ac

R = radius of the circular arc.

cr

 = The angle made by the circular arc at its centre. (a) For circular loop,  = 2 ∴

B=

∴

B=

 ( )  

(× into the page)

On substituting the values, we get B = 6.3 T

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(b) Let the loop is placed in x-z plane. For convenience, we present y-z plane in the plane of paper. We consider a d element at point A on the loop. The co-ordinates of point A are (0, R cos , R sin ). ∴ The position vector of point A is ⃗⃗⃗ = R cos  ̂ + R sin  ̂

But

ee .in

The position vector of point ⃗⃗⃗ = x ̂ dl = R d

Also, current element d = dl cos (90 + ) ̂ + dl sin (90 + ) ̂

itj

= - dl sin  ̂ + dl cos  ̂

ki

The magnetic field at point P due to considered element is

cr ac

For Biot-Savart law d⃗ = =





⃗

 ̂} (⃗⃗⃗⃗ ⃗⃗⃗⃗ )

 ̂

{

|⃗⃗⃗⃗

d⃗ = = =





⃗⃗⃗⃗ | ̂ (

 ̂

*



̂

*

 ̂ (





(

)

{

 ̂|

 ̂ ̂

|

 ̂)

 ̂ ̂

 ̂

 ̂+

)

̂

 ̂

̂+

(∵ dl = R d )

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d Bx =

∴

Bx =

Similarly,

dBy =

∴

By =

Similarly,

Bz =





(

)





(

)

∫





(

)





(

)





(

)

=





(

)

cos  d ∫



=0

∫



=0

∴ Net magnetic field at point P is





(

)

̂

je

⃗ = Bx ̂ + By ̂ + Bz ̂ =

e. in

∴

iit

Hence, magnetic field at point P is along the axis of the loop.

ck

On substituting the value, we get

cr a

B = 2.3 T

Q.11. A current I flows along a thin wire shaped as a regular polygon with n sides which can be inscribed into a circle of radius R. Find the magnetic induction at the centre of the polygon. Analyze the obtained expression at n  . Sol. ∵ The magnetic field at point P is B1 =



(sin 1 + sin 2)

From B, 1 = 2 =

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Also,

cos 1 =

= = cos 2

or

cos =

∴

d = R cos

∴

B1 =





B1 =

. .

/ /

∴

B=



.



ee .in

Since magnetic field due to all sides are in the same direction /=



itj

magnetic field due to circular loop of radius R.

Q.12. Determine magnetic induction at the centre of a rectangular wire frame

ki

whose diagonal is equal to d = 16 cm and the angle between the diagonals is

Sol.

cr ac

equal to  = 30, the current flowing in the frame equals I = 5.0 A.

The magnetic field due to wire AB is equal to that of due to wire CD. ∵ The magnetic field due to wire AD at point O is ⃗⃗⃗⃗ =



(sin 1 + sin 2) ̂

Here

1 = 2 = 75

Also,

cos 75 =

∴

d1 = cos 75

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⃗⃗⃗⃗ =

∴



tan 75 ̂ =



tan 75 ̂

Similarly, magnetic field at O due to wire CD is ⃗⃗⃗⃗ =  tan 75 ̂ The magnetic field at O due to wire BC is ⃗⃗⃗⃗ =



(sin 1’ + sin 2’) ̂

ee .in

1’ = 2’ = 15

Here In Δ BOF1

cos 15 =

∴

OF = cos 15

∴

d2 = cos 15

∴

⃗⃗⃗⃗ =  tan 15 ̂

cr ac

ki

itj

=

From A, the magnetic field ⃗⃗⃗⃗ due to wire DA is same as that of BC. ∴ ∴

⃗⃗⃗⃗ = ⃗⃗⃗⃗

The net magnetic field at point O is ⃗ = ⃗⃗⃗⃗ + ⃗⃗⃗⃗ + ⃗⃗⃗⃗ + ⃗⃗⃗⃗

∴

B= = =







(tan 75 + tan 15) . .







  

/

 

/

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=

 

=

 

(∵  = 30)

On substituting the values, we get B = 0.10 mT Q.13. Determine the magnetic induction of the field at the point O of a loop with current I, whose shape is illustrated (a) in A the radii a and b as well as the angle  are known, (b) in B the radius a and the side b are known.

(i)

Circular arc of radius b:

je

The magnetic field due to this part is

e. in

Sol. Here the wire may be divided into two parts:

( ̂)

A circular arc of radius a making an angle

ac k

(ii)



iit

⃗⃗⃗⃗ =

=(

)

cr

The magnetic field due to this point at O is ⃗⃗⃗⃗ =



(

)( ̂ )

∴ Net magnetic field at point O is ⃗ = ⃗⃗⃗⃗ =2

⃗⃗⃗⃗ 

(

)3 ( ̂ )

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∴

.

B=





/

The magnetic fields due to wires AB and DE are zero. (b) The circuit consists of five: Circular arc: The magnetic field due to this part is ⃗⃗⃗⃗ =

/

( ̂) =

( ̂)

, ⃗⃗⃗⃗ = 0

e. in

But ⃗⃗⃗⃗

.

(straight wires are passing through point O)

The magnetic field due to part (3) at point O ⃗⃗⃗⃗

/ ( ̂) =

.

je

⃗⃗⃗⃗

iit

i.e.

ac k

∴ The net magnetic field ⃗ at point O = ⃗⃗⃗⃗

=

⃗⃗⃗⃗

⃗⃗⃗⃗ √

.

√

( ̂)

⃗⃗⃗⃗ / ( ̂)

cr

Q.14. A current I flows along a lengthy thin-walled tube of radius R with longitudinal slit of width h. Find the induction of the magnetic field inside the tube under the condition h ≪ R. Sol. The magnetic field at a point inside a hollow long tube is zero. The magnetic field due to a long wire is B= The top view of tube. The cutting slit behaves as a long wire. The current in the long slit is

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= ∴ The magnetic field due to slit at point P is ∴

=

=

The magnetic field due to tube is

= 0

⃗ = ⃗⃗⃗⃗

The net magnetic field is

⃗⃗⃗⃗ B=⃗

⃗⃗⃗⃗ =

ee .in

∴

B = | ⃗⃗⃗⃗ | =

∴ ∴

B=

itj

Q.15. Find the magnetic induction of the point O if a current- carrying wire has

ki

the shape A and B. The radius of the covered part of the wire is R, the linear parts

cr

Sol.

ac

are assumed to be very long.

The magnetic fields due to the straight parts at point O are zero. But the magnetic field due to semicircular part is B=

(∵  = )

(b) The given wire may be divided into three parts. (1)

Upper straight wire: This part behaves as semi-infinite wire

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(2)

When, this part is produced left ward, then it is passing through point O. Hence, magnetic field at O due to this part is zero.

(3)

Circular part of radius R: Making an angle  = .

/ at the centre

O. The magnetic field due to this part is 

⃗⃗⃗⃗ = =

/ ( ̂)

. ̂

e. in

=

( ̂)

Q.16. A very long wire carrying a current I = 5.0A is bent at right angles. Find

je

the magnetic induction at a point lying on a perpendicular to the wire, drawn

iit

through the point of bending at a distance l = 35 cm from it.

ac k

Sol. Assume the bending point is at origin. The wire is placed as such one part is along x-axis and other is along z-axis

cr

Both parts of wire behave as semi-infinite wire. The magnetic field at point P due to semi-infinite wire placed along x-axis is

B1 

 

0I ˆ k 4l

The magnetic field at point P due to semi-infinite wire placed along z-axis is

B2 

 

0I ˆ i 4l

The net magnetic field at point P is

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B  B1  B2

B  

0I ˆ 0I ˆ k i 4l 4l 2

 0I   0I    4l  4l 

B=

0I 2 4l

ee .in

=

= 2.0 µT Q.

2

Determine the magnetic induction at the point 0 if the wire carrying a current I = 8.0 A has the shape shown in . a, b, c. The radius of the

The magnetic field due to first semi-infinite wire is

cr

(a)

ki

Solution:

ac

are very long.

itj

curved part of the wire is R = 100 mm, the linear parts of the wire

B1 

 

0I ˆ k 4R

The magnetic field due to semi-circular part is

B2 

 

0I ˆ i 4R

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The magnetic field at O due to third part (semi-infinite wire) is

B3 

 

0I ˆ k 4R

B  B1  B2  B3

B 

 

 

 

0I ˆ  0I  ˆi k 2R 4R 2

2

je

 0I   0I  2R    4R 

iit

B 

0I 4  2 4R

ac k

=

 

e. in

=

 

0I ˆ  0I  ˆi  0I  k ˆ k 4R 4R 4R

cr

On substituting the values, we get B = 0.30 µT.

(b)

The magnetic field at point O due to first part (semiinfinite wire) is

B1 

 

0I ˆ k 4R

The magnetic field at point O due to second part (semicircular) is

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B2 

 

0I ˆ i 4R

The magnetic field at point O due to third part (semi-infinite wire) is

B3 

 

0I ˆ i 4R

B  B1  B2  B3

 

 

iit

0I ˆ k     1 i   4R

0I 2  2  2 4R

ac k

=

 

0I ˆ  0I  ˆi  0I  ˆi k 4R 4R 4R

je

B 

e. in

The net magnetic field at point O is

B 

cr

= 0.34 µT

(c)

The ratio of angle should be equal to ratio of resistances in

the case of uniform wire in circular form. From . D,

 2 R1  I1 R 2

3 = 2 3  2

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 I2 = 3I1 But I1 + I2 = I or 4I1 = I  I1 =

I 4

and I2 =

3I 4

 

0I ˆ k 4R

je

B1 

e. in

The magnetic field due to first part (semi-infinite wire) is

 



0I1  3  ˆ 0I2    ˆ i  i 4R  2  4R  2 

ac k

B2 

iit

The magnetic field due to second part at the centre

 



0I  3  ˆ  3I    ˆ i  0 i   16R  2  16R  2 

cr

 B2 

 B2  0 The magnetic field at point O due to third part (semi-infinite wire) is

B3 

 

0I ˆ j 4R

 net magnetic field at point O is

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B  B1  B2  B3 =

 

0I ˆ  0I   j k 4R 4R

B  B =

0 I 2 4 R

Find the magnitude and direction f the magnetic inductioil vector B of an infinite plane carrying a current of linear density i; the

je

(a)

vector i is the same at all point of the plane:

 0i 2

ac k

B=

iit

 The magnetic field near infinite plane is

directed parallel to the plane.

cr

Q.

e. in

= 0.11 µT

The direction of magnetic field. We now consider a long strip of thickness dx. The considered element behaves as a long wire carry current di = idx where i = Linear current density. The magnetic field due to considered element at point P is dB =

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tan  =

x d

 x = d tan  

dx  dsec2  d

 dx  dsec2  d similarly,

r d

ee .in

sec  

 r  dsec 

itj

di  idx and dx  dsec2 d

ac



0di 2 r

ki

dB =



cr

Putting the values, we get

0idsec2 d dB = 2dsec  =

0i sec d 2

dB = – dB cos  ˆj – dB sin  ˆi

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 Bx  

x 

x  

dB sin 

0I sec  sin d x   2 x 

=



=

0I 2



 /2

 / 2

 tan d  0

 Bx  0



x  



n 

=



 /2

=

0I 2

n 

0I sec  cos d 2

ki

 / 2

dB cos 

ac

  2  2   

0I 2

cr

=

dBy

ee .in

x  

itj

By 

=

 B  Bxˆi  Byˆj =

0I ˆ j 2

B 

0I 2

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(directed parallel to the plane and perpendicular to the linear current density)

A uniform current of density j flows inside an infinite plate of thickness 2d parallel to its surface. Find the magnetic induction induced by this current as a function of the distance x from the median plane of the plate. The magnetic permeability is assumed to

e. in

the equal to unity both inside the outside the plate.

iit

from median plane.

je

Let us consider an infinite plane element of thickness dr at distance r

i=

ac k

The linear charge density in the considered element is

I Jldr  l l

= Jdr

cr

Q.

 The magnetic field at point P due to considered plane is dB =

0i 0 jdr  2 2

(Here, x < d) The net magnetic field at point P is B=

0idr  x 2 x

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=

0 j x r  x 2

=

0 j  x  x 2

when x < d = µ0jx when x < d

d

0 jdr  0 jd 2

iit

A direct current I flows along a lengthy straight wire. From the point 0 the current spreads radially all over an infinite conducting plane

ac k

perpendicular to the wire. Find the magnetic induction at all point of space.

From ampere's circuital law.

cr

Q.



d

je

B=

e. in

when point P is outside the plane, then x > d.

For (1)

 B.dˆi   I 0

c

or B 2r  0I

0I 2r

B  For (2),

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 B.d1  0 C

(net current is zero) B=0 A current I flows along a Tound loop. Find the integral



B dr-along

the axis of the loop within the range from  to + Explain the result obtained.



2 R 2  r2



3 /2

je

B=

0IR 2

e. in

 The magnetic field on the axis of circular loop is

ac k

iit

along the axis

According to situation,



0IR 2

cr

Q.





B dr 

2 R



2

r

0IR 2  Bdr  2 



2



3 /2

dr

dr





R

2

 r2



3 /2

Put r = R tan   dr = R sec2  d when r = ,

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 , 2



when r = – ,

 2

0IR2   Bdr   2 

0I  / 2 cos d 2  / 2

=

0I  /2 sin  / 2  2

iit

= µ0I

je

=

R sec2 d  / 2 R3 sec3   /2

e. in







Q.

Bdr  0I

cr



ac k

After integrating, we get

A direct current of density j flows along a round uniform straight wire with cross-section radius R. Find the magnetic induction vector of this current at the point whose position relative to the axis of the wire is defined by a radius vector r. The magnetic permeability is assumed to be equal to unity throughout all the space.  The magnetic field at a point outside the hollow cylinder is B=

 0I 2 r

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But the magnetic field at a point inside a hollow long cylinder is zero. The top view of round uniform wire. Case-I: When point P is at distance r < R from the axis of the wire: For this, we consider a long hollow cylinder of radius x and thickness dx

e. in

The cross-section area of considered element is dA = 2xdx.

je

The current in the considered element is

iit

dI = J2 xdx

ac k

= 2Jxdx

The magnetic field at point P due to the considered element is

 0dI 2 r

cr

dB =

0 2Jxdx 0 2r r

B 



 J = 0 r

 r2  2  

B 

0 Jr 2





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When r > R, Then B =



R

0

0J2xdx 2r

0J R 2 = r 2 B 



ee .in

Inside a long straight uniform wire of round cross-section there is a long round cylindrical cavity whose axis is parallel to the axis of the wire and displaced from the latter by a distance l. A direct current of density j flows along the wire. Determine the magnetic induction

ki

itj

inside the cavity.

ac

 B  B1  B2

cr

Q.



0R2 Jr 2r2

B1 

0 J  r1 2





B1 

0 J  r1 2





The magnetic field at point P due to cavity is

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B2 



0 J  r2 2



 The net magnetic field is

B  B1  B2

B



0 J  r1  r2 2



From addition law of vectors,

e. in

OC  CP  OP



ac k

Determine the current density as a function of distance r from the axis of a radially symmetrical parallel stream of electrons if the magnetic induction inside the stream varies as B = br, where b and

cr

Q.



0 J l 2

iit

B 

je

or l  r2  r1

 are positive constants. From Ampere's circuital law. According to this law,

 B d1   I 0

C

The current I=

r

 J 2r  dr 0

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 B.dL   I 0

C

or br  r  0

0



or br  1  0 or b

r

 J 2r  dr r

Jrdr

0



r

0

Jrdr

d  1 r  0Jr dr

 

or b    1 r   0Jr

and cross-section radius

ac

A single-layer coil (solenoid) has length

R, A number of, turns per unit length is equal to n. Determine the magnetic induction at the centre of the coil when a current I flows

cr

Q.

ki

itj

b    1 r  1 J  0

ee .in

or B2r  0

through it.

Let us consider a circular coil of radius R and thickness dx. The number of turns in the considered element is dn = ndx. The magnetic field at point P due to considered element is dB =



dn0IR 2

2 R 2  x2



3 /2

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=

0IR 2ndx



2 R 2  x2



3 /2

0IR 2n dx or dB = 2 R 2  x2





3 /2

Its direction is along the axis of solenoid

cot  

x R

1 0nI sin d 2

iit

 dB  

je

 dx   R cosec2 d

e. in

From . A,

=

1 0nI 2

2

   sin  d 1

cr

B=

ac k

The magnetic field at point P due to entire solenoid is

1 0nI  cos 2  cos 1  2

It point P is the at the centre of coil.

cos 1 

1 l2 2 R  4 2

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l

=

4R 2  l2

Similarly,

cos 180  2  

4R 2  l2

0nI  2R  1   l 

2

ki

A very long straight solenoid has a cross-section radius R and n

ac

turns per unit length. A direct current I flows through the solenoid. Suppose that x is the distance from the end of the solenoid, measured along it axis Determine:

cr

Q.

l

itj

B 

4R2  l2

ee .in

 cos 2 

1

(a)

the magnetic induction B on the axis as a function of x;

draw an approximate plot B vs the ratio (b)

x ; R

the distance x0 to the point on the axis at which the value of B differs by  = 1% from that in the middle section of the solenoid.

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 Here B =

0 nI  cos 2  cos 1  2

(along the axis of solenoid)

Solution: (a)

For long solenoid, 1 = 0

R 2  x2

je

x

R

2

 x2



1/2

iit

 cos 2 

x

e. in

and cos (180 – 2) =

0nI cos 2  cos 1  2

ac k B

 0nI  x 1   2  R 2  x2 

cr

=

(b)

According to problem,

B0  B B 1  1  B0 B0 Here B = µ0nI and B0 

0nI  1  x0    2  R 2  x20   

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putting the value of B and B0,

1

 1 x0 1   1  2 2 2 R  x0  

or 

x0 R 2  x20

 1  2





or x20  1  4  42 R2  x20

 5R

je

A thin conducting strip of width h = 2.0 cm is tightly wound in the

iit

shape of a very long coil with cross-section radius R = 2.5 cm to

ac k

make a single-layer straight solenoid. A direct current I = 5.0 A flows through the strip. Determine the magnetic induction inside and outside the solenoid as a function of the distance r from the axis.

cr

Q.

1  2 R 2  1  

e. in

 x0 



Since here, current is flowing in two mutual perpendicular directions. (1)

A part of current I1 is flowing along the length of solenoid.

Due to this current, the system behaves as a hollow long tube of radius R. The magnetic field due to long hollow tube is B1 = 0 at r < R

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B2 =

 0I 2 r

at r > R (2)

Remaining part of current is flowing along the circumference of the coil. Due to this, the system behaves as a long solenoid. The magnetic field due to this is

e. in

B2 = µ0nI2

1 h

iit

Here n =

je

(where r < R)

ac k

(number of turns per unit length)

 B2 

0 I2 h

cr

(when r < R)

 h  Here I2 = I 1    2R 

2

  h   B2  0 I 1    2R  h

2

when r > R, B2 = 0

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Hence, at r < R

B  B1  B2   h  B  0 I 1    2R  h

2

B  B1  B2

B 

itj

N = 2.5.103 wire turns are uniformly wound on a wooden toroidal

ki

core of very small cross-section. A current I flows through the wire.

ac

Determine the ratio  of the magnetic induction inside the core to that at the centre of the toroid.  The magnetic field inside the core is

cr

Q.

0I 2r

ee .in

At r > R

B1 = µ0nI

Here n = number of turns per unit length =

N 2 R

 B1 

0NI 2R

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For calculation of magnetic field at the centre of toroid, the system behaves as circular coil,

 B2 

0I 2R

(at the centre of circular coil)



e. in

A direct current I = 10A flows in a long straight round conductor. Find the magnetic flux through a half of wire's cross-section per one

je

metre of its length.

iit

 Magnetic flux is given by

ac k

   B.dS

The magnetic field at a point inside a long solid wire is

B



0 Jr 2



cr

Q.

B1 N = B2 

0Jr 2

B 

Let us consider an element of width dr at distance r from the axis of wire. The area of considered element is dS = ldr

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(But l = 1 metre)  dS = dr  =



0Jr dr 2

R

0

(Here J =

I ) R 2

ee .in

0J R 2 = 2 2 0IR 2 = 4R 2

itj

A very long straight solenoid carries a current I. The cross-sectional

ac

area of the solenoid is equal to S, the number of turns per unit length is equal to n. Find the flux of the vector B through the end

cr

Q.

 0I 4

ki

=

plane of the solenoid.  At the end of a long solenoid, the magnetic field is B=

 0nI 2

  B.S = BS cos 0

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=

. shows a toroidal solenoid whose cross-section is rectangular. Determine the magnetic flux through this cross-section if the current through the winding equals I = 1.7 A, the total number of turns is I = 1000, the ratio of the outside diameter to the inside one is  = 1.6 and the height is equal to h = 5.0 cm.

e. in

The magnetic field at distance r from centre of toroid is B = µ0nI

 0NI 2r

je

=

iit

Let us consider a strip of thickness dr at distance r from centre O.

ac k

The area of considered element is dS = hdr

 d  B.dS

cr

Q.

0nIS 2

= Bhdr cos 0

 

=



b

a

0NI hdr 2r

0NIh b ln 2 a

 

0NIh ln  2

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Q.

Determine the magnetic moment of a thin round loop with current if the radius of the loop is equal to R = 100 mm and the magnetic induction at its centre is equal to = 6.0 µT.  Magnetic moment of a loop is Pm = NIS Here N = 1, S =  R2

Find the magnetic moment of a thin wire with a current I = 0.8 A,

cr

Q.

ac k

2BR 3  Pm  0

je

2RB 0

iit

or I =

0I 2R

e. in

B 

wound tightly on half a tore. The diameter of the cross-section of the tore is equal to d = 5.0 cm, the number of turns is N = 500.

 Magnetic moment is a vector quantity. Its direction is in the direction of area vector. It is given by Pm = NIS Let us consider an element making angle d at O.

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The number of turns in the considered element is

N d 

dN =

 dPm  dNIS

d2 = dNI 4

 Pmx 

0

d2 dNI sin  4

itj





 0

dPm sin 

ki

=



 0

NId2  cos 0  4

ac

=

 



 

ee .in

dPm  dPm cos  ˆj  dPm sin  ˆi

cr

NId2 = cos   cos 0 4

NId2 = 2 Similarly, Pmy =





0

NId2 cos  d 4

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NId2  = sin 0  0  4  Pm  Pmx

NId2 = 2 = 0.5 A.m2 (in the sense of magnitude)

e. in

A thin insulated wire forms a plane spiral of N = 100 tight turns carrying a current 1 = 8 mA. The radii of inside and outside turns (.)

je

are equal to a = 50 mm and b = 100 mm. Determine: the magnetic induction at the centre of the spiral;

(b)

the magnetic moment of the spiral with a given current,

ac k

iit

(a)

The magnetic field due to a circular coil of N turns at its centre is

B

0NI 2R

cr

Q.

Let us consider a circular element of radius r and thickness dr.  The number of turns in the considered element is dN =

N dr ba

The magnetic field at its centre due to considered element is dB =

dN0I 2r

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B 

=

N0I 2 b  a



b

a

dr r

0NI b ln 2 b  a a

= 7 µT (b)

From the concept of previous problem, dPm = dNIr2



itj

3 b  a

NI 2 b  ab  a2 3



ac

=



NI b2  a3

ki

 Pm 

ee .in

 N  b 2 r dr b  a a

Pm = I  



Q.

cr

= 15 mA.m2

A non-conducting thin disc of radius R charged uniformly over one side with surface density a rotates about its axis with an angular velocity . Find: (a)

the magnetic induction at the centre of the disc;

(b)

the magnetic moment of the disc.

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When a circular ring of radius R is uniformly, charged with the charge q and rotates with constant angular velocity

about its centre. Then

the convection current due to motion of ring is

=

Rr dt

=

q d R 2R dt

I 

e. in

dq dt

q 2

je

I=

Let us consider a ring element of radius r and thickness dr.

cr

(a)

 0I 2R

ac k

B=

iit

The magnetic field at the centre of ring is

The area of considered element is dA = 2rdr  The electric charge on the considered element is dq = adA = 2ardr The convection current due to considered element is

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dI =

=

dq 2

2a rdr 2

= a rdr The magnetic field at the centre due to considered element is

0a r dr 2r

je

=

 0dI 2r

e. in

dB =

iit

 The magnetic field due to disc is

B 

0a 2r

B 

0 a R 2

ac k

R



dr

cr

0

(b)

The magnetic moment due to considered ring is dPm = r2 dI = r2a rdr = a r3dr

 Pm  a



R

0

r3dr

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R4  Pm  a 4

a R 4 = 4 A non-conducting sphere of radius R = 50 mm charged uniformly with surface density  = 10.0 µC/ma rotates with an angular velocity = 70 rad/s about the axis passing through its centre. Determine

e. in

the magnetic induction at the centre of the sphere.



2 R 2  x2



3 /2

iit

B=

0IR 2

je

The magnetic field on the axis of a circular ring is

ac k

Let us consider a circular ring of radius r and thickness dS = Rd So, area of considered ring is dA = 2r dS

cr

Q.

= 2 (R sin ) (Rd) = 2R2 sin  d dq = dA = 2R2 sin  d  dI = =

dq 2

(R2sin  d)

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=  R2 sin  d The magnetic field at centre O of the sphere due to considered element is dB =

0dIr2



2 r 2  x2



3 /2

Here r = R sin  and x = R cos 

0

je

3 /2

 0 R sin3  d 2

iit





ac k

B 



2 R 2 sin2   R 2 cos2  



e. in

 dB 

B 



0 R 2 sin  d R 2 sin2 





0

 0 R sin3  d 2

cr

After integrating, we get B=

2  0 R 3

On substituting the values, we get B = 29 × 10–29 T  20 pt Q.

A charge q is uniformly distributed over the volume of a uniform ball of mass m and radius R which rotates with an angular velocity

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about the axis passing through its centre. Find the respective magnetic moment and its ratio to the mechanical moment. The magnetic moment due to a disc is

q r2 Pm = 4

ee .in

Let us consider a circular disc of radius r and thickness dS = Rd

The volume of considered element is



ac

= R3 sin2 d

ki

=  (R sin )2 (Rd)

itj

dV = r2 dS

r  R sin 

cr

 Electric charge on considered element is dq = R3  sin2 d Here  

=

q 4 R 3 3

3q 4R 3

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

 dq  R3 sin2 d

=

 43q R

3

3q sin2 d 4

The magnetic moment due to considered element is

=

3q sin2 d R 2 sin2  16

je

=

3q sin2 dr2 16

e. in

dPm =

dqr2 4





ac k

3q R2 Pm = 16

iit

 The magnetic moment of the sphere is

0

sin4 d

cr

After integrating, we get

qR 2 Pm = 5

The mechanical moment (angular momentum) of the sphere about its diameter is M=I =

2 mR 2 5

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

A long dielectric cylinder of radius R is statically polarized so that at all its points the polarization is equal to P = r, where is a positive constant, and r is the distance from the axis. The cylinder is set into rotation about its axis with an angular velocity . Find the magnetic induction B at the centre of the cylinder.

e. in

We know that polarisation vector P is numerically equal to surface charge density due to polarisation.

ac k

iit

(at surface, r = R)

je

P = R

Let us consider a ring element of thickness dx at distance x from centre of cylinder.

The charge on considered element is

cr

Q.

Pm q  M 2m

dq = (2Rdx)p = 2R2 dx

 The convection current is dI =

=

dq 2 2R 2dx 2

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R2dx

 dI 

The magnetic field at the centre due to considered element is dB =

0dIR 2



2 R 2  x2



3 /2

The magnetic field at the centre due to whole cylinder is







2 R 2  x2



3 /2

After integrating, we get

je

Putting x = R tan ,

iit

dx = R sec2 d

cr

 2

ac k

when x = + 



e. in

B1 =

0R 2 R 2dx



and x = – ,



 2

0R 4   B1  2 =

0R 4  4R3



 /2

 / 2

R sec2 d  / 2 2R3 sec3   /2

cos d

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=

0R   /2 sin  / 2  4

B1 =

0R  2

Due to polarisation, volume charge density is given by

=

Px Py Pz   x y z

Here P   r

ee .in

   div P

itj

Let axis of cylinder is along x-axis.

ki

Let us consider a point A (x, y, z) in the cylinder.

ac

ˆ  r  y ˆi  zk

ˆ  P  y ˆj  zk



cr



P  r

   

x z   2 x z

Let us consider a circle of radius r and thickness dr having length dx along the axis of cylinder. The current in considered element is

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dI = 2dV

2

= 2 2rdrdx

2

= 2rdr dx The magnetic field due to volume charge density is



R

0

0 r3dr

 0 R 2

1





r

2

2

x



3 /2

dx

ac

=

3 /2

ki

After integrating





ee .in

=



2 r2  x2

itj

 

B2 =

0r2 2r dxdr

cr

 B = B1 + B2

On substituting the values, we get B=0 Q.

Two protons move parallel to each other with an equal velocity v = 300 km/s. Determine the ratio of forces of magnetic and electrical interaction of the protons.  The magnetic force at a point due to moving charge is

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B



0q0 v  r



4r3

The magnetic force on a charge particle of charge q in a magnetic field B is

Fm  qv1  B The magnetic field due to first proton on the site of second proton is



0 e v  r 4r



e. in

B

3

Fm  e v  B



=



ac k

0e v  v  r

iit

je

The magnetic force on second proton is

4r3

0 2 v r 4r3

cr

=

The electrical force of interaction between them is

e2 r Fe  40r3 

Fm Fe

 v200

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But c =

1 00

 00 



Fm Fe

1 c2

v2  v   2    c c

2

Fm

iit

Determine the magnitude and direction of a force vector acting on a

ac k

unit length of a thin wire, carrying a current I = 8.0 A, at a point 0, if the wire is bent as shown in (a)

with curvature radius R = 10 cm;

cr

Q.

 106

je

Fe

e. in

On substituting the values, we get

(b)

the distance between the long parallel segments of the wire being equal to

= 20 cm.

The magnetic force on a wire is given by

Fm  Id L  B  Force per unit length is

Fm IdLB sin   dL dL

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= IB sin  (a)

The magnetic force at point O due to semi circular wire is

B

 

0I ˆ k 4R

Here 

 

0I ˆ k 4R

itj

dL  dL ˆi

ee .in

B 

ki

 F  Id L  B

 

 

IdL0I ˆ j 4R

cr =

 

0I ˆ k 4R

ac

i  = IdL  ˆ

 

F 0I2 ˆ   j dL 4R



F I  02 dL 4R

On substituting the values, we get

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F  0.2 mN / m dL (b)

Find magnetic field at point O: The magnetic field at point O due to upper semi-infinite wire

is

 

0I ˆ k L 4    2

e. in

B1 

The magnetic field at point O due to lower semi-infinite wire

 

iit

0I ˆ k L 4    2

ac k

B2 

je

is

 Net magnetic field at point O is

cr

    I 0I  ˆ 0 B   k L L      4   4     2     2 

=

 

 

0I ˆ k l

Here dL  dL ˆj

 Fm  IdL  B

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

 

I ˆ = IdL ˆj  0  k L

0I2dL ˆ = k L Fm 0I2   dL L On substituting the values, we get

e. in

Fm  0.13  103 N / m dL = 0.13 mN/m

je

A copper wire with cross-sectional area S = 2.5 mm2 bent to make

iit

three sides of a square can turn about a horizontal axis 00'. The wire

ac k

is located in uniform vertical magnetic field. Determine the magnetic induction if on passing a current I = 16 A through the wire the latter deflects by an angle  = 20°.

cr

Q.

In this case, magnetic torque about OO' is balanced by torque due to gravitational torque. If OO' is connected by a wire, then the torque due to magnetic force on this wire about OO' is zero. So, for solving the problem, the system may be assumed as a square loop. The area of loop is A = l2 At angle of deflection the magnetic torque is

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B  Pm  B Here Pm = magnetic moment of loop = IA = Il2 The angle between normal to plane (area vector or magnetic moment) and magnetic field is 90° – B = Il2 B sin (90° – 0) = Il2 B cos 

e. in

The mass of each side is m = volume × density = (Sl) lS

je

The torque due to weight of wires is

ac k

= 2mglsin

iit

1 1 g  mg sin   mg sin   mglsin  2 2

cr

= 2Sl2 gsin  For equilibrium,

B  g  Il2B cos   2Sl2gsin 

B 

2Sg tan  I

= 10 mT

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Q.

A small coil C with N = 200 turns is mounted on one end of a balance beam and introduced between the poles of electro-ciagnet as shown in . The cross-sectional area of the coil S = 1.0 cm2, the length of the arm OA of the balance beam is

= 30 cm. When there is no

current in the coil the balance is in equilibrium. On passing a current I = 22 mA through the coil the equilibrium is restored by putting the additional counterweight of mass m = 60 tng on the balance pan.

e. in

Find the magnetic induction at the spot where the coil is located.

Solution:

ac k

First:

iit

je

This problem can be solved in following ways:

Discuss the problem in the absence of current in the coil.

cr

When current in the coil is zero, then, torque due to magnetic force is zero. In this case, the torque of weight of coil about point O is balanced by the weight of the balance.

 m1gl1  mgl

…(i)

Now discuss the problem, when current is passing through the coil. In this case, torque due to magnetic force is

1  NISB Anticlockwise

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The torque due to weight of coil is 1 = m1gl1 The torque due to the weight of the balance + additional weight m about O is 3 = (m + m) gl And for equilibrium,

1  2  3

e. in

 m1gl1  NISB  m  m gl

(from eq. (i)

mgl NIS

ac k

B 

iit

or NISB  mgl

je

or m1gl1  NISB  mgl  mgl

cr

On substituting the values, we get B = 0.4 tesla. Q.

A square frame carrying a current I = 0.90 A is located in the same plane as a long straight wire carrying a current I0 = 5.0 A. The frame side has a length a = 8.0 cm. The axis of the frame passing through the midpoints of opposite sides is parallel to the wire and is separated from it by the distance which is  = 1.5 times greater than the side of the frame. Determine:

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(a)

Ampere force acting on the frame;

(b)

the mechanical work to he performed in order to turn the frame through 180° about its axis with the currents maintains if constant.

 The magnetic field at a point due to a long wire is B=

0I0 2r

e. in

(a)

The magnetic force on wire section (1) and (3) are in

iit

 F1   F3

je

opposite directions and are equal in magnitude.

ac k

The magnetic field at the site of wire (2) due to long wire is

0I0

ˆ k  a

 2  a    2

cr

B1 

The magnetic force on wire section (2) is

F2  IaB1 ˆi = Ia

0I0

a  2  a    2

ˆi

Similarly,

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F4 

 

Ia0I0  ˆi a  2  a    2

Similarly,

F4 

 

Ia0I0  ˆi a  2  a    2

e. in

 Net force on the frame is

F  F1  F2  F3  F4





je

20ii0 ˆ i  42  1

iit

=

ac k

On substituting the values, we get F = 0.4 × 10–6 N

cr

= 0.4 µN

(b)



d dt

or power P = I = I

d dt

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work done by magnetic force is A = Pdt = – Id

 A   I  work done by external agent is Aext = – A

Aext  I

e. in

= I  f  i 

je

We consider an element of thickness dx

iit

 d  B.dS

0I0 dx 2x

ac k =  Badx  

cr

I  i   0 0 2

 i  

a 2 a a 2



a

dx x

0I0  2  1 ln   2  1 2

when frame is turned through 180°.

f  i =

0I0a  2  1 ln   2  1 2

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   f  i =

0I0a  2  1 ln   2  1 

 Aext  I =

0II0a  2  1 ln   2  1 

e. in

On substituting the values we get, = 0.10 µJ

je

Two long parallel wires of negligible resistance are connected at one end to a resistance R and at the other end Wage source. The

iit

distance between the axes of the wires is  = 20 times greater than

ac k

the crass-sectional radius of each wire. At what value of resistance R does the resultant force of interaction between the wires turn into zero?

cr

Q.

The system formed by given long parallel wires is a capacitor.

C  l

20  b  a ln   a 

where l is length of each long wire. or C =

20l b ln a

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 =

=

b  a

20l a ln a 20l ln 

This problem can be solved by following ways:

e. in

First:

Draw equivalent circuit of the problem.

iit

q 0 C

ac k

V0 

je

In loop (1)

 q  CV0

20l V0 ln 

cr

=

 

=

q l

20V0 ln 

…(i)

In loop (2),

q  IR  0 C

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I 

q V  0 RC R

…(ii)

q   V0   C

 

Now discuss the problem on the basis of force: In this case, two forces of interaction between wires comes into play. (i)

Electrical force of interaction:

e. in

The electric field due to positive plate at the site of negative

 20r

iit

E=

je

plate is

q 20lr

=

q 20la

cr

ac k

=

The electrical force of interaction is Fe = qE =

 l  l 20l a

2l = 20 a

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(ii)

Magnetic force of interaction: From Ampere's law of magnetic force, two long parallel wires carrying current in opposite directions repel each other with a magnetic force Fm =

0I1I2l 2r

r = a

0I2l 2 a

je

 Fm 

e. in

Here I1 = I2 = I,

iit

Since, resultant force of interaction is zero.

ac k

Hence, Fm = Fe

cr

0I2l 2l   2a 20 a

2 or 0I  0 2

…(iii)

From Eqn. (i) and (ii), we get I=

V0 R

and  

20V0 ln 

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From Eqn. (iii), we obtain

2 0I  0 2

Putting the value of i and  in eqn. (iii), we get R = R0

0 ln  0 

e. in

=

On substituting the values, we get

A direct current I flows in a long straight conductor whose cross-

ac k

section has the form of a thin half-ring of radius R. The same current flows in the opposite direction along a thin conductor located on the "axis" of the first conductor (point O in .). Determine the

cr

Q.

iit

= 0.36

je

R0 = 0.36 × 103

magnetic interaction force between the given conductors reduced to a unit of their length. The magnetic field due to a long straight conductor whose crosssection is in the form of a thin half ring of radius R is B=

 0I 2R

The magnetic force on a straight wire of length dl at point O is

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dE = Idl B sin 90° = Idl

=

0I 2R

0I2dl 2R

 Force per unit length is

Two long thin parallel conductors of the shape shown in . carry direct currents I1 and I2. The separation between the conductors is

je

a, the width of the right-hand conductor is equal to b. With both

iit

conductors lying in one plane, find the magnetic interaction force

ac k

between them reduced to a unit of their length. From Ampere's law of magnetic force, two long parallel wires carrying current in same direction attract each other with a force

cr

Q.

e. in

dF 0I2  2 dl R

F 0I1I2  l 2r

Let us consider a long strip of thickness dr at distance r from first wire. The current in considered element is dI =

I2 dr b

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The magnetic force of interaction between first wire and considered element is dF =

=

0I1dI 2r

0I1I2 dr 2b r

a

dr r

0I1I2 a  b ln 2b a

ac k

A system consists of two parallel planes carrying currents producing a uniform magnetic field of induction B between the planes. Outside this space there is no magnetic field. Find the magnetic

cr

Q.



a b

iit

F

0I1I2 2b

je

F 

e. in

(per unit length)

force acting per unit area of each lane. Since, magnetic field outside the planes is zero. Hence, linear current density j in both planes are opposite to each other. Here magnetic field at point P situated between the planes is B = B1 + B2 =

0i 0i   0i 2 2

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i 

B 0

The magnetic field due to 1st plane at the site of second plane is B1 =

 0i 2

Now we consider an element of length l and breadth b in the second plane. The current in the considered element is I2 = ib

e. in

 magnetic force on considered element F = I2lB

je

 0i 2

iit

= (ib) l

ac k

Force per unit area is

F 0i2  lb 2

2

cr

B 0    0  = 2

B2 = 2 0 Q.

A conducting current-carrying plane is placed in an external uniform magnetic field. As a result, the magnetic induction becomes equal to B1 on one side of the plane and to B2, on the other. Find magnetic

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force acting per unit area of the plane in the cases liustrated in . Determine the direction of the current in the plane in each case. When magnetic lines of force in a region is denser, magnetic field is stronger. (a)

From of problem,

B1 > B2 The magnetic field due to plane is

 0i 2

ee .in

Bplane =

Here i is linear current density

0i 2

itj

 B1  Bex 

ki

…(i)

0i 2

ac

 Bex  B1 

cr

Similarly,

B2  Bex 

0i 2

or B2 = B1 

0i 0i  2 2

i 

…(ii)

B1  B2 0

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Bex  B1  = B1 

=

2  B1  B2  2  0 

B1 B2  2 2

B1  B2 2

 The magnetic force on an element of length l and breadth

e. in

b is F = IlBex

je

= (ib) lBex

iit

Force per unit area is

ac k

F  iBex lb

 B1  B2   2 

cr

= i

=

B1  B2  0

 B1  B2    2 

B12  B22 = 20 (b)

From of problem, B1 > B2

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also, Bex < Bplane Here B1 = Bex  Also B2 

0i 2

0i  Bex 2

Bex 

B1  B2 2

and i =

B1  B2 0

ee .in

After solving,

ac

ki

F  iBex ib

itj

Force per unit length is

cr

B12  B22 = 20

 Force per unit area is

F  iBex sin  lb B12  B22 = 20

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(c)

From , it is clear that Bex and Bplane are inclined to each other shown in . 3.260C. Let Bex is dire

-

axis.



B12  Bplane  Bex sin 

  B 2

cos 

2

ex



and B22  Bex cos  Bex sin   Bplane 2



= 4Bex Bplane sin

2

  B 2

ex sin   Bplane

e. in

 B12  B22  Bplane  Bex sin 



je

  i  B12  B22  4  0  Bex sin   2

ac k

iit

= 20 iBex sin 

cr

B12  B22  Bex sin   20 Force per unit area is

F  iBex sin  lb

B12  B22 = 20 Q.

In an electromagnetic pump designed for transferring molten metals a pipe section with metal is located in a uniform magnetic field of induction B. A current I is made to flow across this pipe

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section in the direction perpendicular both to the vector B and to the axis of the pipe. Determine the gauge pressure produced by the pump if B = 0.10 T, 1 = 100 A, and a = 2.0 cm.

Now we consider a strip of length l and thickness dx. The current in the strip is

I dx. a

e. in

I1 =

The area of strip is

je

dA = Idx

ac k

dF = I1lB

iit

The magnetic force on the considered element is

I  dx lB a 

cr

= 

Pressure is P=

=

dF dA

I lBdx a ldx

The pressure is P=

IB a

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= 0.5 kPa. A current I flows in a long thin-walled cylinder of radius R. What pressure do the walls of the cylinder experience? Let us consider a long strip of thickness (Rd) carrying a current dI =

Id 2

The magnetic field at point P due to considered element is

=

0Id 42r

itj

0 Id 2r 2

ac

=

ee .in

 0dI 2 r

ki

dB =

2 +  = 180°

 2

cr

Q.

   90 

R 2  R 2  r2 Also, cos   2R2

 r2  2R2 1  cos  2 2 = 4R sin

 2

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 r  2 sin  dB 

=

 2

0Id   42  2R sin   2

0I d 82R sin

 2

 Bx  

 2 

 0

ee .in

 dB  dB sin  ˆi  dB cos  ˆj dB sin 

0I 2 cot d 82R 0

cr

=

ac

ki

itj

  0I sin  90   2  2 =  d 0  2 8 R sin 2

After integrating, Bx = 0 Similarly, By = 



 2 

 0

dB cos 

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  0I cos  90   2  2 = d 0  2 8 R sin 2

0I 2 d 82R 0

=

0I 2 82R

=

0I 4R

e. in

=

je

The magnetic force on a strip of thickness b and length l is

I I bl  0 2R 4R

cr

F 

I b 2R

ac k

Here I2 =

iit

F = I2 l By

 Pressure =

F area

=

F l b

0I2lb = 82R 2

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0I2 = 82R 2 Q.

What pressure does the lateral surface of a long straight solenoid with n turns per unit length experience when a current I flows through it? The average magnetic field due to long solenoid is

1 0nI 2

The magnetic force is

0n2I2lb 2

ac

=

1 0nI 2

ki

= nbIl

itj

F = (nbI)lB

ee .in

B=

cr

Pressure =

F lb

0n2I2 = 2 Q.

A current I flows in a long single-layer solenoid with cross-sectional radius R. The number of turns per unit length of the solenoid equals

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n. Find the limiting current at which the winding may rupture if the tensile strength of the wire is equal to Flim. From previous problem, the force on an element l is dF =

1 0nI2 l 2

For equilibrium,

1 0nI2 l 2

or T 

1 0nI2R 2

2T 0nR

cr

I 

iit

l  R

ac k



je

or T 

e. in

T  dF

 Ilim 

=

 Q.

2Tlim 0nR

2Flim 0nR Tlim  Flim 

A parallel-plate capacitor with area of each plate equal to S and the separation between them to d is put into stream of conducting

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liquid with resistivity . The liquid moves parallel to the plates with a constant velocity v. The whole system is located in uniform magnetic field of induction B, vector B being parallel to the plate and perpendicular to the stream direction. The capacitor plates are interconnected by means of an e eternal resistance R. What amount of power is generated in that resistance? At what value of R is the generated power the highest? What is this highest power equal to?

e. in

The system behaves as a parallel combination of a capacitor and resistor. Conducting liquid consists of positive and negative ions. Since, system is placed in an external magnetic field B. So, motional

iit

Solution:

je

emf is generated between plates of capacitor.

ac k

The equivalent circuit of the problem is For equilibrium of ion,

cr

eE = evB

 E = vB

Motional emf is s = Ed = vBd The electric resistance between plates of capacitor is

r

d S

Also capacity is

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C=

0S d

In loop ABCDA,  – Ir – IR = 0

 vBd  r R r R

or I 

vBd d  R S

ee .in

I 

The power generated in external resistance R is

2

itj

P = I 2R

ac

ki

   Bvd  P =   R d  R   S

cr

For maximum power generated,

dP 0 dR

After solving, we get R=

S d

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 Pmax

B2v2d2  S      2S   d    d 

B2v2d2 = 2S d v2B2Sd = 4

e. in

A wire bent as a parabola y = ax2 is located in a uniform magnetic field of induction B, the vector B being perpendicular to the plane x,

je

y. At the moment t = 0 a connector starts sliding translation wise

iit

from the parabola apex with a constant acceleration w. Determine the emf of electromagnetic induction in the loop thus formed as a

ac k

function of y.





The motional emf produced in a rod is = B. 1  v . If B, l and v are

cr

Q.

mutually perpendicular. Then induced emf in the rod is  = Blv Now the problem can be solved in following ways. Find the effected length of wire: The length of rod in time t is l = 2x Here y = ax2

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x  l 

y a y a

But v2 = 2wy

 v  2wy Find induced emf

je

8w a

A rectangular loop of with a sliding connector of length

is located

in a uniform magnetic field perpendicular to the loop plane. The

cr

Q.

2wy

ac k

= By

y a

iit

= B2

e. in

  Blv

magnetic induction is equal to B. The connector has an electric resistance R, the sides AB and CD have resistances R1 and R2 respectively. Neglecting the self-inductance of the loop determine the current flowing in the connector during its motion with a constant velocity.

This problem can be solved by following ways: Find direction and magnitude of induced emf.

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The movable rod behaves as a real battery of emf  = Blv and internal resistance R Now draw equivalent circuit of the problem: =

Bvl R  R

The circuit of the problem is

  IR  I1R1  0

In loop (2)

iit

From KVL

…(i)

je

or Blv  IR  I1R1  0

e. in

In loop (1),

ac k

  IR  I  I1  R2  0

…(ii)

cr

After solving eq. (i) and (ii), we get I=

=

=

 R1R 2 R R1  R 2

Bvl R1R 2 R R1  R 2

Bvl R  R

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 R1R 2  where R    R1  R2  A metal disc of radius a = 25 cm rotates with a constant angular velocity

= 130 rad/s about its axis. Determine the potential

difference between the centre and the rim of the disc if (a)

the external magnetic field is absent;

(b)

the external uniform magnetic field of induction B = 5.0 ml

e. in

is directed perpendicular to the disc.

We a particle move on a circular path with constant speed, radial

je

acceleration is acted on the particle radially inwards.

(a)

2

ac k

wn = r

iit

The radial or centripetal acceleration is given by

When conducting disc is in rotating motion with constant angular velocity

about its axis. Every electron moves on a

cr

Q.

circular path with same angular velocity . Let us consider an electron at distance r from centre of disc.

 eE  mwn

mr E  e

2

(where m is mass of electron) But d  Edr

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2

 d  1



a

0

mr e

2

dr

After integrating, we get

m 2a2 1  2  2e On putting the values, we get

1  2  3nV In the presence of magnetic force, the motional emf in an

e. in

(b)

d = B (dr) v

je

element of thickness dr at distance r from centre is



a

B rdr

ac k

 

iit

= B(dr) (r ) = B rdr

0

cr

B a2 = 2

On substituting values, we get

1  2   = 20 × 10–3 V = 20 mV Q.

A thin wire AC shaped as a semi-circle of diameter d = 20 cm rotates with a constant angular velocity = 100 rad/s in a uniform

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magnetic field of induction B = 5.0 mT, with

|| B. The rotation

axis passes through the end A of the wire and is perpendicular to the diameter AC. Determine the value of a line integral

 E dr along

the wire from point A to point C. Generate the obtained result.

Let us consider a dL = Rd  element at an angle  with diameter. The motional emf in the considered element is



ee .in



d= B. dL  v

= BdLv sin 

ki

= B (Rd) (r ) sin 

itj

= B.  dL sin   ˆ n

ac

 d  BRr sin d

cr

From the . A,

    2  

 

 2

  d  BRr sin d 2

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R 2  R 2  r2 Also cos      2R2  2R2  r2  2R2 cos 

 r2  2R2 1  cos  2 2 = 2R 2 cos

 2

(from i)





iit

2R  cos 2 sin 2 d

ac k

 d  BR

 2

je

 r  2R cos

e. in

2 2 = 4R cos

 2

cr

= BR2 sin d

or d  BR2 sin d

   BR2 = BR2





0

sin d

  cos 



0

   2BR2 C

  Edr   A

= 2BR2

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C

  Edr  2BR2 A

 d Edr   2B   A 2 C

=

2

1 B d2 2

= – 10 mV

A wire loop enclosing a semi-circle of radius a is located on the

je

boundary of a uniform magnetic field of induction B. At the moment t = 0 the loop is set into rotation with a constant angular

iit

acceleration about an axis 0 coinciding with a line of vector B on

ac k

the boundary. Determine the emf induced in the loop as a function of time t. Draw the approximate plot of this function. The arrow in the shows the emf direction taken to the positive.

cr

Q.

d  2R 

e. in



In this case, induced emf is generated due to change in area inside the magnetic field. The area of loop in side the magnetic field B in time t is

R2  A= 2 The flux passing through loop is

  BA

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  =B

d dt

dA dt

BR 2 d = 2 dt



1 2 t 2

e. in

But  

d  t dt

iit

BR 2t 2

ac k

=

je

BR 2   t 2

cr

when loop enters in the magnetic field current is anti-clockwise. But when loop is coming out from the magnetic field, current is in clockwise direction. In general,

BR2t    1 2 n

Ba2Bt    1 2 2



R  a

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where n number of half revolution.

A square frame with side a and a long straight wire carrying a current I are located in the same plane. The frame translates to the right with a constant velocity v. Determine the emf induced in the frame as a function of distance x.



d dt

je

(According to Faraday's law)

e. in

The induced emf in a closed loop is

iit

This problem can be solved by following ways:

ac k

Find net flux in the loop:

Let us consider an element of width dr of the loop. The magnetic field due to the long wire in the considered element is

cr

Q.

B=

 0I 2 r

The magnetic flux in the considered element is

d  Badr =

0Ia dr 2r

 

0Ia x  a dr 2 x r

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=

0Ia  a ln 1    2 x



d dt

 induced emf

d dt

 

d  0Ia  a  ln 1      dt  2 x 

cr

ac k

0Ia2v = a  2x2 1    x

je

0Ia a dx a  x2 dt  2 1    x

iit

=

e. in



0 2Ia2v = 4x  x  a Q.

A metal rod of mass m can rotate about a horizontal axis O, sliding along a circular conductor of radius a. The siiangement is located in a uniform magnetic field of induction B directed perpendicular to the ring plane. The axis and the ring are connected to an emf source to form a circuit of resistance R. Neglecting the friction, circuit inductance, and ring resistance, find the law according to which the

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source emf must vary to make the rod rotate with a constant angular velocity . This problem can be solved by following ways: Find determine motional emf in the rod. We consider an element of length dx at distance x from the centre O of the ring. The motional emf in considered element is

e. in

d = B(dx) v

= B xdx

je

= B(dx) x



a

0

B xdx

1 2 Ba 2

cr

=

=

ac k

in

iit

Net motional emf in the rod is

Now analyze the mechanical condition of rod: Since, rod is rotating about point O with constant angular velocity  So, the net torque on the rod about point O is zero. For this, torque due to magnetic force on rod is balanced by torque due to gravitational force on the rod.

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The torque due to gravitational force on rod about point O is

a g  mg sin t 2 (clockwise direction) For balancing gravitational torque by torque due to magnetic force. The magnetic torque should be anticlockwise. This is possible only when the direction of current is from O to A in the rod. We consider

e. in

an element dx of rod. The magnetic force on the considered element is

iit

je

dF = I(dx)B

ac k

The torque due to this force df about O is d g = x dF

cr

(anti-clockwise direction) or d B = IBxdx a

 B  IB xdx 0

IBa2 = 2 (anti-clockwise) For equilibrium of rod,

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B  g IBa2 a or  mg asin t 2 2

I 

mgsin t aB

…(i)

Make equivalent circuit of the system:

–in – IR = 0 in + IR

Q.



a2B2  2Rmgsin t



aB

cr

1 = 2

iit

je

1 2 Rmgsin t Ba  2 aB

ac k

=

e. in

The equivalent circuit is shown. According to loop rule,

A copper connector of mass m slides down two smooth copper bars, set at angle  to the horizontal, due to gravity. At the top the bars are interconnected through a resistance R? The separation between the bars is equal to

. The system is located in a uniform magnetic

field of induction B, perpendicular to the plane in which the connector slides. The resistances of the bars, the connector or and the sliding contacts, as well as the self-inductance of the loop, are

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assumed to be negligible. Find the steady-state velocity of the connector. For steady state velocity of rod, magnetic force on the rod is balanced by component of weight of rod along the inclined plane. On the basis of conservation principle of energy, the power supplied by weight of rod appears as thermal power in the circuit. This problem can be solved by following ways:

e. in

First:

Determine motional emf in rod at steady state.

je

Let the velocity of rod at steady state is v0.

iit

The motional emf in the rod is = Blv0

ac k

Now draw the equivalent circuit of the system. According to loop rule,

cr

 – IR = 0

I 

=

 R

Blv0 R

Now discuss the system on basis of energy conservation principle. The power supplied by weight of the rod is

Pg  mg.v0

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= mgv0 sin The thermal power developed in the circuit is P = I 2R 2

 Blv0  = R  R 

B2l2v20 = R

Pg = P

The system differs from the one examined in the foregoing problem

cr

Q.

mgR sin  B2l2

ac k

 v0 

iit

je

B2l2v20  mgv0 sin   R

e. in

From principle of conservation of energy,

by a capacitor of capacitance C replacing the resistance R. Determine the acceleration of the connector. At t = 0, the velocity of rod is zero. But acceleration of rod is w=

mgsin  m

= g sin  where m is mass of rod.

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when t > 0, motional emf is founed into the rod which is cause of flow of charge through the circuit. Let at an instant t, current in the circuit is i. According to loop rode,



q 0 C

q = C

dv  BlCw dt

ac

ki

BlC

dq dt

itj

i 

ee .in

= BlCv

It means current in circuit depends upon acceleration of the rod.

cr

From free diagram of rod . mg sin  – ilB = mw or mg sin  – (Bl) BlCw = mw or mg sin = (m + B2l2C) w

w 

mgsin  m  B2l2C

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=

A wire shaped as a semi-circle of radius rotates about an axis OO' with an angular velocity

in a uniform magnetic field of induction B

The rotation axis is perpendicular to the field direction. The total resistance of the circuit is equal to R. Neglecting the magnetic field of the induced current, determine the mean amount of thermal

d dt

iit



je

The induced emf in a loop is

e. in

power being generated in the loop during a rotation period.

First:

ac k

This problem can be solved in following ways:

Determine induced emf in the loop.

cr

Q.

g sin  l2B2C 1 m

Let the area of rectangular part of circuit is A0. The magnetic flux through rectangular part ABCD of the circuit is 1 = BA0 At t = 0, the area vector of semicircular part, and vector B are in the same direction. At an instant t, the angle between area vector and vector B is  = t (i.e. the angular displacement of semicircular part in tune t)

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So, the magnetic flux through semicircular part of the loop at an instant t is

a2 2  B cos  2 Ba2 cos t 2 The net flux through the loop at an instant t is

B2 cos t 2

je

= BA0 

e. in

2

BdA0 Ba2 d  cos t dt 2 dt

cr

=

d dt

ac k



iit

The induced emf in the loop is

Ba2 sin t =0 2 Now draw equivalent circuit of the system. The equivalent circuit of the loop is shown

Ba2  sin t 2 According to loop rule;

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I=

 R

Ba2 sin t = 2R The instantaneous thermal power in the circuit is P = I 2R

2

sin2 t 4R

je

=

2B2a4



T

2

iit

Mean power is

Pdt

0

ac k

 P 

e. in

2B2a4 2 sin2 t R = 4R2



t

2

dt

0

cr

2

2B2a4 = 4R

2



0

sin2 t 2

After integrating, we get,

 = Q.

a2B



2

2R

A small coil is introduced between the poles of an electromagnet so that its axis coincides with the magnetic field direction. The cross-

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sectional area of the coil is equal to S = 3.0 mm2, the number of turns is N = 60. When the coil turns through 180° about its diameter a ballistic galvanometer connected to the coil indicates a charge q = 4.5 µC flowing through it. Determine the magnetic induction magnitude between the poles provided the total resistance of the electric circuit equal R = 40 .

IR =

Nd dt dq d N dt dt

or R

q  N t t

cr

ac k

or R

je

Nd dt

iit



e. in

According to Faraday's Law of electromagnetic induction:

or Rq  N

 q 

=

N R

N f  i R

Solution: The initial flux through the coil in i  BS

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when coil is rotated through 180°, the area vector and magnetic field are in opposite direction.  Flux through coil after 180° rotation of coil is

f  B.S = BS cos 180° = – BS

B 

qR 2NS

je

2NBS R

ac k

q=

N f  i R

iit

 q 

e. in

 f  i  2BS

cr

On substituting the values, we get B = 0.5 tesla Q.

A square wire frame with side  and a straight conductor carrying a constant current I are located in the same plane The inductance and the resistance of the frame are equal to L and R respectively. The frame was turned through 180° about the axis OO' separated from the current carrying conductor by a distance b. Determine the electric charge having flown through the frame.

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During rotation of loop external flux (flux due to magnetic field of long straight wire) crossing through the loop changes. This change in external magnetic flux is caused by induced emf in the loop. This induced emf behaves as external voltage source for the loop. Due to this emf, current starts to flow in the circuit. When current starts to flow in the circuit, Self induced emf is generated in the circuit which opposes the flow of current. In nut

ee .in

shell the system behaves as L — R circuit in series. The problem is solved in following steps: First:

ki

itj

Draw the equivalent circuit at an instant t:

ac

In the shown circuit,

cr

 = induced emf due to change in external magnetic flux. ' = Self induced emf in the loop According to loop rule, – i R – ' = 0 (At an instant t) or

d di  Ri  L 0 dt dt

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or

d di  Ri  L 0 dt dt

or

d dq di R L 0 dt dt dt

or

 q i R L 0 t t t

or   Rq  Li  0 Here i  if  ii

ee .in

…(i)

Since, initial current in the circuit is zero

 ii  0

itj

After 180° rotation, finally flux becomes constant. Hence, final

ac

If = 0

ki

current in the circuit is

cr

 i  if  ii  0

From Eq (i), we get  – Rq – 0 = 0

 q 

 R

…(ii)

Now determine net change in external flux:

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At the initial position of loop (frame), magnetic field and area vector are opposite in direction. But after rotation of frame through 180°, the area vector and vector B becomes in the same

   2   1  = 2  1

   2  1



r b  a

=



ba

=

0ai  b  a ln   b  a 2

je

0i adr 2r

iit

ba

BdA

e. in

r b  a

ac k

=

From Eq. (ii), we get

=

= Q.

cr

q = q

 R

0aI  b  a ln 2R  b  a

A conducting rod AB of mass m slides without friction over two long conducting rails separated by a distance

(.). At the left end the

rails are interconnected by a resistance R. The system is located in a uniform magnetic field perpendicular to the plane of the loop. At

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the moment t = 0 the rod AB starts moving to the right with an initial velocity v0. Neglecting the resistances of the rails and the rod AB, as well as the self-inductance, determine: (a)

the distance covered by the rod until it comes to a

standstill; (b)

the amount of heat generated in the resistance R during

e. in

this process.

At t = 0, when rod start to move, motional emf is produced in the

je

rod. Due to this a current starts to flow through circuit. According to Lenz's law, the direction of current is as such the magnetic force on

iit

the rod opposes the, motion of rod. It means magnetic force on the

ac k

rod is in opposite direction of velocity of rod. Due to this, rod starts to decelerate and finally comes in rest. In the reference of energy

cr

conservation principle, total kinetic energy of rod appears of thermal energy in the circuit during the process. (a)

The motional emf in rod at an instant t is  = Blv where v is speed of rod at an insant t. The current in the circuit at an instant t is I=

 R

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=

Blv R

The magnetic force on the root at an instant t is F = IlB

B2l2v = R The deceleration in rod is

F m

e. in

w=

je

B2l2v = mR

ac k

iit

dv B2l2 or v  v dS mR

cr

But w = v

or 



0

v0

S  (b)

dv ds

dv 

B2l2 S dS mR 0

mRv0 B2l2

Thermal energy generated in the circuit = loss in kinetic

energy of rod =

1 mv20 2

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A connector AB can slide without friction along a U-shaped conductor located in a horizontal plane (.). The connector has a length

, mass

m, and resistance R. The whole system is located in a uniform magnetic field of induction B directed vertically. At the moment t = 0 a constant horizontal force F starts acting on the connector shifting it translationwise to the right. Determine how the velocity of the connector varies with time t. The inductance of the loop and the

e. in

resistance of the U-shaped conductor are assumed to the negligible.

At t = 0, the velocity of rod is zero. So, no motional emf is induced in

F at t = 0. Due to this m

iit

je

the rod. But rod has acceleration w 

ac k

acceleration, rod starts to accelerate in forward direction. When t > 0, speed of rod is greater than zero. So, motional emf is generated in the which is equal in the rod is  = Blv. Due to this emf, current will be in the circuit. Let current in the circuit

cr

Q.

at instant t is i a velocity of the rod is v.

The equivalent circuit is shown According to loop rule,  – IR = 0 Blv – IR = 0

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I 

Blv R

…(i)

Due to this current, a magnetic force acts on the rod in the opposite direction of applied force F. The force-diagram of rod is shown

w

F  ilB m

v

ki

mdv  B2l2v F R



t

cr



ac

Blv    i  R 

itj

B2l2v dv F  R or  dt m

0

ee .in

The acceleration of rod at an instant t is

0

dt

After solving we get,

v

F 1  et m





B2l2 Here   mR

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Q.

. illustrates plane s made of thin conductors which are located in a uniform magnetic field directed away from a reader beyond the plane of the drawing. The magnetic induction starts diminishing. Determine the induced in these loops are directed. This problem is based upon Lenz's law and is an example of statically induced 6mf.

e. in

The direction of statically induced emf is determined on the basis of magnetic field and the direction of dynamically induced emf is

A plane spiral with a great number N of turns wound tightly to one

iit

another is located in a uniform magnetic field perpendicular to the

ac k

spiral's plane. The outside radius of the radius of the spiral's turns is equal to a. The magnetic induction varies with time as B = B0 sin t, where B0 and

are constants. Determine the amplitude of emf

cr

Q.

je

determined on the basis of magnetic force.

induced in the spiral. Let us consider an element of radius r and thickness dr

 dN 

N dr a

 The magnetic flux in considered element is d= Br2 The induced emf in considered element is

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d = – dN

= dN r2

d dt

dB dt

= dNr2 B0

sin t

Total emf generated is

N 2 r dr 0 a

t

a

1 B0 Na2 cos t 3

1 B0 Na2 3

ac k

 max 

A U-shaped conductor is located in a uniform magnetic field

cr

Q.

r 0

dNr2

iit

=

sin

r a

je

= B0

t

e. in

 = B0 sin

perpendicular to the plane of the conductor and varying with time at the rate B = 0.10 T/s. A conducting connector starts moving with an acceleration w = 10 cm/s2 along the parallel bars of the conductor. The length of the connector is equal to

= 20 cm.

Determine the emf induced in the loop t = 2.0 s after the beginning of the proton, if at the moment t = 0 the loop area and the magnetic induction are equal to zero. The inductance of the loop is to be neglected.

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Here emf is induced due to change in area of the loop as well as the change in magnetic field.

 

d dt

d BA  dt

=

BdA dB A dt dt

ee .in

=

Now the distance travelled by the connected loop in time t is

1 2 wt 2

itj

x

ki

The area of the loop at an instant t is

ac

F = xl

 1 2 wt  l  2

cr

= 

wlt2 = 2

The magnetic field in the loop at instant t is B = Bt



dB B dt

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  B

dA dB A dt dt

d  wlt2  wlt2 = Bt  B dt  2  2

wlt2 = Btlwt  B 2 Blwt2 = Blwt  2 =

e. in

2

3 Blwt2 2

iit

ac k

 = 12 × 10–3 V

je

On substituting the values, we get

= 12 mV

In a long straight solenoid with cross-sectional radius a and number of turns per unit length n a current varies with a constant velocity I

cr

Q.

A/s. Determine the magnitude of the eddy current field strength as a function of the distance r from the solenoid axis. The magnetic field due to solenoid is B = µ0nI The induced emf in the loop is = 

d dt

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 = Ba2 = µ 0 n I  a2 = µ0 na2I

d dt

2 = 0na

dI dt

= 0a2nI



or  E.dr  0na2I

je

C

e. in

  

iit

(for r > a)

ac k

or E2r  0na2I

cr

0na2I E  2r when r < a,

 E.dL   C

d dt

2 or E  2r  r

E 

dB dt

r d  0nI 2 dt

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0nrdl 2dt

=

0nr I 2

A long straight solenoid of cross-sectional diameter d = 5 cm and with n = 20 turns per one cm of its length has a round turn of copper wire of cross-sectional area S = 1.0 mm2 tightly put on its winding. Find the current flowing in the turn if the current in the solenoid

ee .in

winding is increased with a constant velocity I = 100 A/s. The inductance of the turn is to be neglected.

d dt

ac

Here   BA

ki



itj

From Faraday's law of electron magnetic induction,

= 0nIr2

cr

Q.

=

d2 0nI 4

0nId2 = 4

 

d dt

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d2 dI = 0n 4 dt = 0n

d2 I 4

l S

=

2 S

d S

ac k

=

iit

 d 2    2 = S

je

R=

e. in

The resistance is

cr

 The induced current is I=

 R

0nd2 I 4 = d S =

0nISd 4

On substituting the values, we get

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I = 2 × 10–3 A = 2 mA A long solenoid of cross-sectional radius a has a thin insulated wire ring tightly put on its winding; one half of the ring has the resistance  times that of the other half. The magnetic induction produced by the solenoid varies with time as B = bt, where b is a constant. Determine the magnitude of the electric field strength in the ring.

d Ba2 dt





2 = a

ac k

(Here B = bt)

je

=

d dt

iit



e. in

The motional emf in the ring is

dB dt

cr

Q.

=  a2 b

…(i)

Since, induced emf does not depend upon resistance, so, in whole ring induced emf is same. But q 

 q 

 R

1 R

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From this point of view, transfer of charge depends upon resistance. It means the current will be different in two parts. For symmetry of current, electric field is induced, which produces emf in both parts in opposite directions. For half part of ring having resistance R0, a    E.dL  IR 0 0 2 a    EdL cos180  IR 0 0 2

or

  Ea  IR 0 2

e. in

or

je

…(ii)

ac k

  Ea  IR 0 2

iit

Similarly, For second half part,

…(iii)

cr

From equ. (i), (ii) and (iii), we get

   1 ab

E=     1 2 Q.

A thin non-conducting ring of mass m carrying a charge q can-freely rotate about its axis. At the Initial moment the ring was at rest and no magnetic field was present. Then a practically uniform field was switched on, which was perpendicular to the plane of the ring and increased with time according to a certain law B (t). Determine the angular velocity co of the ring as. A function of the induction B(t).

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2  E.dL  r C

2 or E2r  r

dB dt

dB dt

r dB 2 dt

E 

…(i)

1 dB  r  2  dt 

je

E

e. in

In vector

ac k

 = qEr

iit

The torque due to electric force qE about point O is

or mr2 = qEr

cr

( = angular acceleration)

qr2 dB or mr   2 dt 2

(from equ. (i)) or  

q dB 2m dt

But  

d dt

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

d q dB  . dt 2m dt

or





q B t  dB 2m 0

q B t 2m

or







q B t 2m

e. in

0

d

(In vector from)

A thin wire ring of radius a and resistance r is located inside a long

je

solenoid so that their axes coincide. The length of the solenoid is

iit

equal to , its cross-sectional radius, is b. At a certain moment the

ac k

solenoid was connected to a source of a constant voltage V. The total resistance of the circuit is equal to R. Assuming the inductance of the ring to be negligible. Determine the maximum value of the radial force acting per unit length of the ring.

cr

Q.

The problem can be solved by following ways: Now determine self-inductance of solenoid: As we know, N = LI N(BA) = LI or N  0nIA   LI

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



or N 0nIb2  LI

 L  0Nnb2 = 0 nl nb2 = 0n2b2l Now draw equivalent circuit of solenoid:

e. in

According to loop rule,

V L

=



t

0

dI 0 V  IR dt

iit

I

ac k

or L

dI  IR  0 dt

je

V – ' – IR = 0

cr

After solving, we get I=

V 1  eRt /L R





…(i)

Now find the induced current in the ring: The flux associated with ring is  = Ba2 = 0nIa2

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Here B = magnetic field due to solenoid = 0nI …(ii) Induced emf in the ring is

dr dt

2 = 0na

dI dt

….(ii)

The current in the ring is

…(iii)

ac k

0na2 dI  = r dt

je

' r

iit

I' =

e. in

' 

From equ. (i), we get

V 1  eRt /L R



cr

I





dI V Rt /L  e dt L

0na2 V Rt /L  I'  e r L (from equ. (iii)) Now find magnetic force on an element of length dl of the ring: The magnetic force on the considered element is

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dF = I'dLB. Force per unit length is

F0 

dF  I'B dL

On putting the value we get,

0na2V Rt /L F0 = e  0nI Lr 20n2a2V Rt /L V e 1  eRt /L Lr r





e. in

=

20n2V2 Rt /L  F0  e 1  eRT /L LrR



iit

je



ac k

For maximum value of F0,

dF0 0 dt

cr

After solving, we get

 dF   dL  max

F0max =  

20a2V2 = 4rRlb2 Q.

A particle with specific charge q/m moves in the region of space where there are uniform mutually perpendicular electric and magnetic fields with strength E and induction B. At the moment t =

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0 the particle was located at the point O and had zero velocity. For the non-relativistic case determine: (a)

the law of motion x (t) and y (t) of the particle; the shape of the trajectory;

(b)

the length of the segment of the trajectory between two nearest points at which the velocity of the particle turns into zero; the mean value of the particle's velocity vector projection

e. in

(c)

on the x axis (the drift velocity).

At t = 0, the charged particle is at rest. So, magnetic force is

je

(a)

zero but electric force acts along y-axis, which accelerates

iit

the charged particle along y-axis. When t > 0, charged

ac k

particle picks up speed, variable magnetic force starts to act which has a tendency to make curve the path of particle back

cr

around towards x-axis. Let the velocity of charged particle at point P is

v  vxˆi  vyˆj So, Lorentz force on the charged particle at point P is

F  qE  qv  B





ˆ ˆ  q vxˆi  vyˆj  Bk or F  qEj

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ˆ  qvxBj ˆ  qvyBiˆ or F  qEj Fx = qvy B

 wx 

qB vy m

…(i)

Also, Fy  qE  qvxB

dwy dt

qB wx m

dwy qB  qB   vy dt m  m 

ac k

or

qB dvx m dt

iit

=



je

or

qE  qvxB m

e. in

or wy 

cr

(from equ. (i))

dwy

q2B2  vy or dt m2 or

d2vy dt2

q2B2   2 vy m

…(ii)

In the case of simple harmonic motion,

y or w  

2

y

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or

dv  dt

2

d2y or  dt2

y

2

y

 y  A sin  t  

…(iii)

The equation (ii) is similar to equation (iii) comparing Eqn. (ii) and (iii), we get

e. in

qB m



…(iv)

je

 vy  A sin  t  

iit

According to problem, at

ac k

t=0

wx = 0,

qE , m

cr

xy =

vx = 0, vy = 0 From equ. (iv) vy = A sin ( t + ) or 0 = A sin ( × 0 + )

0

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 vy  A sin t or

dvy

 A cos t

dt

But at t = 0,



qE m

qE A m qE m

A 

ee .in

wy =

qE sin t m

ki

itj

 vy 

qE sin t qB m m

cr

ac

or vy 

E sin t B

 vy  y  =



t

0

…(v)

vydt

E 1  cos t B

 where

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

qB m

…(iv)

For x-component:

qBvy

wx =

m

or wx =



vx

0

dvx 

e. in

qE t sin tdt m 0

ac k

or

dvx qE  sin t dt m

je

or

qE sin t m

iit

=

qB E sin t m B

qE 1  cos t m

cr

 vx 

or vx 

E 1  cos t B

 

qB  m 

x 





t

0

vxdt t

E  sin t  t =   B  0

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x 

E sin t  t   B 

x 

E B

x 

mE  t  sin t qB2



t  sin t 

mE qB2



qB m

iit

and

je

Here a =

…(vii)

e. in

x = a( t – sin t)

Discussion of trajectory:

ac k

E B



t  sin t  …(viii)

E 1  cos t …(ix) B

cr

x 

and y =

or Y =

mE 1  cos t  qB2

 



qB  m 

or y = a (1 – cos t) From equ. (viii)

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sin t = t –

B x E

From equ. (ix) cos t = 1 

B y E

 sin2 t  cos2 t  1 2

2

2

E  B2 2   x  t  2 B E 2

2

E    y   1 B 

je

B2 2 or E2

e. in

B x  B y  or   t    1  1  E   E 

2

ac

ki it

E  E  E2   or  x  t   y    2 2 …(x)   B  B  B

cr

This is an equation of circle of radius R=

E B

whose centre moves with constant velocity v0 =

E B

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The actual path of particle is cycloid defined as the path generated by a point on the circumference of a circle of radius R =

E . B

The motion of charged particle is similar to a point on the rim of a wheel of radius R rolling with constant speed v0 along x-axis

e. in

It is clear that maximum displacement of charged particle along the y-axis is equal to the diameter of rolling circle.

 From discussion, after one time periodic time, velocity

iit

(b)

je

2E    i.e.  2R  B 

ac k

becomes zero.

qB m

cr



T   

2m qB



2  T 

The instantaneous speed of charged particle is v=

v2x  v2y

Substituting the the values of vx and vv we get

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E 2  2 cos t B

v=

E 2 1  cos t  B

=

E t 2 sin B 2

=

2E t sin B 2

s 



vdt

2  2 m  qB

2E t Sm B 2

itj

0

8mE qB2

ac

s  (c)

0

ki

=

t



T

ee .in

=

From shown in A, drift speed is equal to speed of centre of

cr

the circle

i.e. = Q.

E B

Magnetron is a device consisting of a filament of radius  and a coaxial cylindrical anode of radius b which are located in a uniform magnetic field parallel to the filament. An accelerating potential difference V is applied between the filament and the anode.

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Determine the value of magnetic induction at which the electrons leaving the filament with zero velocity reach the anode. When electron is accelerated through a potential difference V.  eV =

1 mv2 2

v 

2eV m

…(i)

e. in

The magnetic field maintains circular path for electron of radius r. This circle touches the anode.

je

From,

ac k

or b2 – a2 = 2br

iit

a2 + r2 = (b – r)2

…(ii)

cr

b2  a2 r = 2b

From circular motion evB =

mv2 r

r=

mv eB

…(iii)

From equation (i), (ii) and (iii), we get



B=  

2b  2mV  b  a2  e 2

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241

cr

ac ki

itj ee .in

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242