Crackiitjee.in.Phy.ch22

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AMPERE'S LAW We know that the lines of magnetic field are continuous and do not arise from any source in the way as lines of electric field originate from charges. These thus form loops without any beginning or end. This property may be used to go through the nature of magnetic field and for calculating the field in certain situation of high symmetry. The field of a long around current carrying wire B =

0 i . The field 2 r

lines are concentric around the wire in the plane of the paper if the wire

ee .in

is perpendicular to this plane. The line integral around the path of radius r, starting at any point and returning to the same point is;

=

d

cr ac ki

=B

 B d cos 0

itj

 B.d

0 i .  2r 2 r

= 0i.

Although the above result it derived for the special case of the field of a long straight conductor, the law is true for conductors and path of any shape. Thus the line integral of the magnetic field B around any closed path is equal to  0 times the net current across the area bounded by the path. Hence

 B.d

 0 iin

It is called Ampere's law. It plays the same role in magnetic as Gauss's law plays in electrostatics. For more Study Material and Question Bank visit www.crackiitjee.in

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NOTE: 1.

The magnetic field B on the left hand side in Ampere's law is the resultant field due to all the currents existing anywhere while on the right hand side the current iin is due to the conductors inside the close loop.

2.

Let us consider a closed plane curve as shown in figure. If director of integration is taken along the path as shown then i1 and i3 will be positive and i2 will be negative.

.in

Thus the total current crossing the loop is (i1 – i2 + i3). Any current

equation. Thus

Consider two current carrying conductors, carrying currents i 1 and i2 in

ac ki

3.

 0 i1  i2  i3 

itj

 B.d

ee

outside the area is not included in writing the right hand side of the

path. For the close path

 0 i1  i2 

cr

 B.d

Application of Ampere's Law

In some cases of practical importance, symmetry considerations make it possible to use Ampere's law to compute the magnetic field caused by a certain current carrying conductor. Few guiding principles, analogous to those stated for Gauss's law are following. If B is tangent to the integration path everywhere and has the same magnitude on each point of the path, then

B  0i. Here

is the length of the close path. For more Study Material and Question Bank visit www.crackiitjee.in

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If B is everywhere perpendicular to the path or some portion of the path, then that portion of the path makes no contribution to the line integral. In the integral

 B.d

, B is always the resultant magnetic field at each

point of the path. In general this field is caused partly by currents linked by the path and partly by currents outside. Even when no current is linked by the path the field at points on the path need not be zero. In case, however, OB.d is always zero.

ee .in

Some judgement is required in choosing an integration path. Two useful guiding principles are that the point or points at which the field is to be determined must lie on the path, and that the path must have enough

itj

symmetry so that the integral can be evaluated.

cr ac ki

Magnetic field due to long cylindrical wire For r > R:

The magnetic field at each point of circular path surrounded the conductor is tangential. Therefore for circular path 1.

 B.d or

 0i

 Bd

or B

d

cos 0  0i  0i

or B  2r  0i or B =

0 i 2 r

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If the current is distributed uniformly through the cross-section of the wire, then the current inside the path is

i ir2 2 iin =  r  2 R2 R Now

 B.d

 0iin

ir2 or B  2r  0  2 R

0 ir 2 R 2

.in

or B =

0 i . 2 R

itj

Bmax =

ee

For r = R

ac ki

Example. Each of the eight conductors carries 2.0 A of current into or out of the page. Two paths are indicated for the line integral

 B.d

. what is the

cr

value of the integral for the path (a) at the left and (b) at the right? Solution: (a)

By Ampere law

 B.d

 0 iin;

in close loop iin = (– 2 + 2 – 2) A = – 2A



 B.d

 0  2

=  20. Ans. (b)

In this loop iin For more Study Material and Question Bank visit www.crackiitjee.in

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= (2 – 2 + 2 – 2) = 0,

 B.d



 0 Ans.

Example. If current density in the conducting wire is proportional to the distance r from the axis of the conductor, then find magnetic field at the position r < R, where R is the radius of cross section of the conductor. Solution: Let current density j = kr, where r is the distance from the distance from the axis of the wire. If current in the wire is i, then

 j  2rdr

i=

0

R



= kr  2 rdr

itj

0

ee .in

R

cr ac ki

2kR 3 = 3

And current inside the close path of radius r r

iin =

 j  2rdr 0

r



= kr  2 rdr 0

2kr3 ir3 =  3 3 R For r < R:

 B.d

 0iin For more Study Material and Question Bank visit www.crackiitjee.in

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 ir3  or B  2r  0  3  , R  which gives

0i r2 Ans. B= 2 R 3 It is more useful to write Ampere's law as follows:

 B.d

 0  j.d A

.in

Thin walled hallow current carrying tube For x > R,

ee

0 i . 2 r

itj

Bmax =

iin = 0,

 B  0.

ac ki

For x < R,

cr

Example. A cross-section of a large metal sheet carrying an electric current along its axis. The current in a strip of width d is kd where k is constant. Find the magnetic field at a point P at a distance x from the metal sheet. Solution: Consider two strips R an S of the sheet situated symmetrically on he two sides of P. The magnetic field at P above sheet and below sheet is parallel to sheet as shown in figure. There is not field perpendicular to the sheet. Now applying Ampere's law to the close path 1-2-3-4-1;

 B.d

 0iin For more Study Material and Question Bank visit www.crackiitjee.in

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2

 B.d 1

3

4

1

2

3

4

  B.d   B.d   B.d

= 0 k



or B  0  B  0 = 0 k or B =

 0k . Ans. 2

Example. A cross-section of a long thin ribbon of width b that it carrying a

ee .in

uniformly distributed total current i into the page. Calculate the magnitude and direction of the magnetic field B at a point P in the plane of the ribbon at a distance 'a' from its edge.

itj

Solution: Take a small element of thickness dx at a distance x from P, the current

di =

cr ac ki

in the element

i dx, b

and magnetic field at P, dB =

0 di . 2 x

idx  = 0. b 2 x The total field

 i B= 0 . 2 b

 a b



a

dx x

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=

0 i  a  b . ln x a 2 b

=

0 i a  b Ans. . ln 2 b a

Long Solenoid For the field at point inside the solenoid, let us consider a rectangular path ABCDA as the path of integration. This path is particularly simple as: (a)

Outside the solenoid, along CD, we can assume field to be zero as it is

.in

many times weaker than that inside the solenoid D

  B.d  0

Along BC and DA, B is either zero (for outside parts of BC and DA) or

itj

(b)

ee

C

C



A

 B.dl 

(c)

 B.dl  0

D

cr

B

ac ki

perpendicular to B (for the parts inside the solenoid)

Inside the solenoid B is constant and is along AB. B



 B.dl  Bl A

Now apply Ampere’s Law for closed loop ABCDA, we get

 B.dl   i

0 in

B



C



D



A



or B.dl  B.dl  B.dl  B.dl  0iin A

B

C

D

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As iin  nl i  B  0ni

Example: Consider the current carrying loop formed of radial line and segments of circles whose centres are at point P. Find the magnitude and direction of B at point P. Solution: Magnetic field by straight parts of the loop is zero because point P lies on their axis. The field produced by cured parts is

B

60 1   n2 360 6 0i  1 1   12  a b 

itj

As n1 

ee

0n1i 0n2i  2a 2a

ac ki



.in

B  Ba  Bb

cr

out of the page. Ans.

Example: A loop, carrying a current i, lying in the plane of the paper, is in the field of a long straight wire with constant current i0 (inward). Find the torque acting on the loop. Solution : The field due to current carrying wire is tangential to every point on the circular portion of the loop and hence the forces acting on these segments are zero. Now consider two small elements of length dr at a distance r from the axis symmetrically. The magnitude of the force experienced by each element is For more Study Material and Question Bank visit www.crackiitjee.in

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dF = B i dr

 i    0 . 0  idr  2 r  On element 1 it is into the page and on 2 it is out of the page  d  dF  2r sin 

 i i    0 0 dr  2r sin   2r  Now total torque, b

ee

0i0i sin  b  a Ans. 

itj



.in

 i i sin    00 a dr 

ac ki

Example: Eight wires cut the page perpendicularly at the points. A wire labeled with the integer k (k = 1, 2, ….. 8) carries the current k i0. For those

cr

with odd k, the current is out of the page; for those with even k, it is into the page. Evaluate

 B.dl along the closed path in the direction shown.

Solution: In accordance with Ampere's law

 B.dl   i  0

m

= 50i0 Ans. Example: Electrons emitted with negligible speed from an electron gun are accelerated through a potential difference V along the x-axis. These electrons emerges from a narrow hole into a uniform magnetic field B For more Study Material and Question Bank visit www.crackiitjee.in

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directed along this axis. However, some of the electrons emerging from the hole make slightly divergent angles. Show that these paraxial electrons are refocused on the x-axis at a distance x 

82mV . eB2

Solution: The velocity of electrons as they emerge out,

1 mv2  eV 2 2eV m

.in

or v 

ee

The velocity can be resolved into two perpendicular components,

itj

v cos  along x-axis and v sin  to it. The time taken to complete a

T

2m eB

ac ki

circle

cr

In this time the distance travelled by electrons along x-axis

x  v cos   T

 2eV 2m   cos  eB   m for small ,cos   1 x

82 mV Proved eB2

Example: A particle of mass m and charge q is projected into a region having a perpendicular magnetic field B. Find the angle of the particle as it For more Study Material and Question Bank visit www.crackiitjee.in

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comes out of the magnetic field if the width d of the region is very slightly smaller than (a)

mv qB

(c)

2mv qB

(b)

Solution: The radius of path r 

mv 2qB

mv qB

 mv  d  r,  qB 

cr ac ki

(a)

d r

itj

sin  

ee .in

For d  r,we have

 mv   qB  1  sin    mv   qB  or  

(b)

 radian Ans. 2

 mv  d  r,  2qB 

  B0t, mv 2qB 1   mv 2 qB For more Study Material and Question Bank visit www.crackiitjee.in

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or  

(c)

 radian Ans. 6

 2mv  d  r,  qB  the deviation of particle is therefore

   radian. Ans. (iv)

Velocity of particle: We have,   B0t,

v0 B0

ee .in

r

cr ac ki

v  vxˆi  vyˆj

itj

Velocity of particle at any time t,

 v0 cos ˆ i  v0 sin ˆ J

On substituting the value of , we have or v  v0 cos B0t  ˆ i  v0 sin B0t  ˆj (v)

Position of particle:

ˆ r  xˆi  yj = r sin ˆ i  r  r cos  ˆj

 r sin  ˆi  1  cos  ˆj   

v0  sin B0t  ˆi  1  cos B0t  ˆj  B0 





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Example: When a proton is released from rest in a room, it starts with an initial acceleration a0 towards west. When it is projected towards north with a speed of v0 , it moves with an initial acceleration 3a0 towards west. Find the electric field and the maximum possible magnetic field in the room. Solution : When proton released, it experiences only electric force, so

Fe  ma0  Eq

ma0 ma0  , towards west Ans. q e

ee .in

E

and so

Fe  Fm  m 3a0 

cr ac ki

As Fe  ma0 ,

itj

When proton is projected, it experiences force due to both the fields,

 Fm  2ma0.

The magnetic force on the proton will be ev0B, therefore

ev0B  2ma0, B

2ma0 downward Ans. ev0

Comparison of paths in E -field and B -field 1.

  0

1.   0

Fig.

Fig.

Straight line path

Straight line path

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2.

3.

  90

2.   90

Fig.

Fig.

Parabolic path

Circular path

0    90

3. 0    90

Fig.

Fig.

Parabolic path

Helical path

ee .in

Deviation of charged particle in magnetic field

Suppose a charged particle q enters normally in a uniform magnetic field

B . The magnetic field extends to a distance x, which is less than or equal

The radius of path

r

cr ac ki

(i)

itj

to radius of the path, that is x  r

mv qB

and sin  

x r

Above relation can be used when x  r (ii)

For x > r

r

mv qB

and deviation,   180 as clear from the diagram. (iii)

If particle moves for time t inside the field, then

  t For more Study Material and Question Bank visit www.crackiitjee.in

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 Bq  t  m As   q / m    B t

CYCLOTRON -

ee .in

particle, deuteron etc. It is based on the fast that the electric field accelerates a charged particle and the magnetic field keeps it revolving in circular orbits of increasing radius.

itj

It consists of two hollow D-shaped metallic chambers D 1 and D2 called

cr ac ki

dees. The dees are connected to the source of high frequency electric field. The whole apparatus is placed between the two poles of a strong electromagnet N–S. The magnetic field acts perpendicular to the plane of the dees. (i)

Cyclotron Frequency: Time taken by charged particle to describe a semicircular path, t=

 m  v qB

The period of oscillating electric field T = 2t 

2m qB

and cyclotron frequency For more Study Material and Question Bank visit www.crackiitjee.in

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f= (ii)

1 Bq  T 2m

Maximum kinetic energy of the particle: (a) We have

mv qB

for r0 =

mv0 qB

 v0 

r0qB m

itj

1 mv22 2

ac ki

 K.E. =

ee

(r0, maximum radius of circular path)

.in

r=

2

 q2B2  2 1  r0qB  = m  r0  m  2  2m 

cr

(b) Also, K.E., = work done by electric source K.E. = f × [qV × 2] = 2fqV. Example. Two positive ions having the same charge q but different masses, m 1 and m2 are accelerated horizontally from rest through a potential difference V. They then enter a region where there is uniform field B normal to the plane of the trajectory. (a)

Show that, if the beam entered the magnetic field along x-axis, the value of the y-coordinate for each ion at any time t is approximately For more Study Material and Question Bank visit www.crackiitjee.in

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1

 q 2 y = Bx2   8mV  Provided x remain much smaller than y. (b)

Can this arrangement be used for isotope separation?

Solution: The path of ion is shown in fig. 5.23. The magnetic force provides an acceleration which is towards the centre of path C. At any instant the

F qvB  m m

and ay = ac cos  for small x,

itj

ac =

ee .in

acceleration

cr ac ki

(a)

  0, cos   q  ay

ac 

qvB m

For x = vt, y=

…(i)

1 ay t2 2

…(ii)

From above equations, we get

1  x y = ay   2  y =

2

1  qvB  x2 2  m  v2 For more Study Material and Question Bank visit www.crackiitjee.in

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=

1 qBx 2 2 mv

…(iii)

Since ions are accelerated through potential V,



1 mv2  qV 2

Which gives

2qV m

.in

v=

Example 3. A hypothetical magnetic field existing in a region is given by

ee

B  B0er , where er denotes the unit vector along the radial

itj

direction. A circular loop of radius a, carrying a current i, is placed with

ac ki

its plane parallel t the xy-plane and the centre at (0, 0, d). Find the magnitude of the magnetic force acting on the loop.

a

cr

Solution :.

sin  

a2  d2

The given field can be resolved into two components: B 0 sin , which acts all round the periphery of loop and lie in the place of the loop and other component B0 cos  acts all round the periphery and perpendicular to the plane of the loop. Force on the loop due to B2 cos  is zero. The force is only due to B 0 sin , which is F = B0 sin  i  2a For more Study Material and Question Bank visit www.crackiitjee.in

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= 2Bia 

or F =

a a2  d2

2Bia2 a2  d2

Ans.

Example. A circular wire-loop of radius a, carrying a current i, placed in a perpendicular magnetic field B. Take a small part d of the wire. Find forced on this part of

(a)

the wire exerted by the magnetic field. Find the force of compression in the wire.

.in

(b) Solution:

ee

The force on the element by magnetic field dF = Bi  d

(a)

,

itj

towards the centre of loop (By FLHR).

Let us now consider an element, cutout from the loop. The force

ac ki

(b)

exerted by rest part of loop on the element. Let this force is T.

cr

Since net force on the entire loop is zero, so the net force on its element also be zero. Therefore the equilibrium of element gives

2T sin

  Bi d 2

for small , sin

 2T 

d



 2

 2

  Bi  a 2

 a

which gives For more Study Material and Question Bank visit www.crackiitjee.in

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T = Bia Ans. Example. The magnetic field existing in a region is given by

x  B  B0 1   k. 1  A square loop of edge

and carrying a current i, is placed with its

edges parallel to x-y axis. Find the Magnitude of the net magnetic force experienced by the loop. Solution: Magnetic field at x = 0,

ee .in

ˆ B1  B0k and at x  ,

itj

B2  2B0k

cr ac ki

The forces on the sides BC and DA are equal and opposite and get cancelled.

1 Now y = 2

qBx2 2qB m m 1

q 2 2  = Bx   8mV  (b) From equation (iii)

m



1 y

m1 y  2. m2 y1 For more Study Material and Question Bank visit www.crackiitjee.in

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Example. A proton accelerated by a potential difference V = 500 kV flies through a uniform transverse magnetic field with induction B = 0.51 T. The field occupies a region of space d = 10 cm thickness (see fig. 5.27). Find the angle through which the proton deviates from the initial direction of motion.

Solution : If v is the speed of the proton, then

2qV . m

or v =

ee .in

1 mv2  qV 2

cr ac ki

itj

Here q and m are the charge and mass of the proton respectively.

The proton traverses circular path of radius R in the magnetic field, where R=

mv qB

From the figure

sin  

=

d R

d  2qV  m   m  qB

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= Bd

q 2mV

On substituting the values, we get   30. Ans. Example. A particle with specific charge q/m moves rectilinearly due to an electric field E = E0 – ax, where a is a positive constant, x is the distance from the point where the particle was initially at rest. Find (a)

the distance covered by the particles till the moment it come to a standstill; the acceleration of the particle at that moment.

ee .in

(b)

Solution: The force exerted by the electric field on the particle F = Eq = q(E0 – ax).

dv q  E  ax dx m 0

cr ac ki

v

itj

Acceleration of the particle

…(i)

Upon integrating, we have

1 2 1 1  v  E0x  ax2   C   2 m 2 Given at x = 0, v = 0;

 C  0. 2 Thus v 

2q  1  E0x  ax2  …(ii)   m 2

(a) The velocity of the particle again will be zero. 0=

2q  1 2 E x  ax   m  0 2 For more Study Material and Question Bank visit www.crackiitjee.in

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or x = xm 

2E0 . Ans. a

(b) The corresponding acceleration, from equation (i) =

2E  q E0  a  0   m a 

= 

qE0 Ans. m

Example. A slightly divergent beam of non-relativistic charged particles accelerated by a potential difference V propagates from a point A long

distance

ee .in

the axis of a straight solenoid. The beam is brought into focus at a from the point A at two successive values of magnetic

induction B1 and B2. Find the specific charge q/m of the particles.

itj

Solution:

cr ac ki

Suppose v0 is the velocity of the particle, then

1 mv20  qV 2 or v0 

2qV m

…(i)

For slightly divergent particle,   0

 v0 cos 

v0.

If t is the time to travel a distance , then t =

v0

. In this duration, let

particle completes n circles for B 1 and (n + 1) for B2, then

v0

n

2m qB1 For more Study Material and Question Bank visit www.crackiitjee.in

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= n  1

2m qB2

…(ii)

On solving n=

B1 B2  B1

Substituting this values of v0 and n in equation (ii) and solving, we get

q or  m

82V

2

B2  B1 

2

Ans.

ee .in

 B1  2m   B2  B1  qB1 2qV m

Solution: (a)

cr ac ki

shown in the figure.

itj

Example. Find the force on the conductors placed in uniform magnetic field as

F = Bi (PQ sin ),

direction of force along +z-axis (b) PQ = 0 F = Bi × 0 = 0 The force on side AB,

F1  B0i j And the force on side CD,

F2   2B0i j For more Study Material and Question Bank visit www.crackiitjee.in

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Therefore net force on the loop

F  F1  F2 =  B0i j or F = Bi Ans.

POTENTIAL ENERGY OF MAGNETIC DIPOLE

.in

When the current loop or a magnetic dipole is placed in magnetic field

ee

B, work must be done by an external agent to change the orientation of the dipole. Thus work done by agent becomes the potential energy of

itj

the magnetic dipole. Magnetic energy is assumed to be zero when M is

ac ki

 to B. If the dipole is rotated through an angle position, then



=



cr

U = Wagnet

 d

90



= MB

 sin  d

90

or U = – MB cos  or U =  M

B

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Above results are derived for rectangular loop. but they can be used for a plane loop of any shape. 1.

For circular loop: M = iA = iR2 and  = MB sin  (iR2) B sin  Fnet = 0 For solenoid : IF windings are closely spaced, the solenoid can be

.in

2.

itj

 = NiAB sin 

ee

approximated by a number of circular loops.

ac ki

Example. Four orientations, at an angle , of a magnetic dipole moment M in a magnetic field. Find :

(a) the magnitude of the torque on the dipole and

Solution :

cr

(b) the potential energy of the dipole.

For the dipole 1. 1   2. 2  180   3. 3  180   4. 4  360   (a)

Torque on the dipole is given by  = MB sin . For more Study Material and Question Bank visit www.crackiitjee.in

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1 = MB sin  2 = MB sin (180° – ) = MB sin  3 = MB sin (180° + ) = – MB sin  4 = MB sin (360° – ) = – MB sin 

U = – MB cos . U1 = – MB cos  U2 = – MB cos (180° – )

cr ac ki

= MB cos

ee .in

Potential energy of a dipole in magnetic field is given by

itj

(b)

U3 = – MB cos (180° + ) = MB cos 

U4 = – MB cos (360° – ) = – MB cos

Analogy between electric dipole and magnetic dipole

1.

Electric dipole

Magnetic dipole

Pq

1.

direction from

M  iA direction from S to N

– q to + q 2.

Fnet  0

2.

Fnet  0

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3.

  P E

3.

  MB

4.

U = – PE cos 

4.

U = – MB cos

More about M (a)

The direction of M is the direction of thumb of the right hand if the fingers of this curl around the loop are in the direction of the current.

(b)

Direction of M can also be determined as the case of electric dipole, the dipole moment P has a direction from negative to positive charge. In the similar way direction of M is from south magnetic pole to north

.in

magnetic pole. The south the north poles can be determined by the

ee

sence of flow of current. The side from where the current seems to be clockwise becomes south pole [S] and the opposite side from it seems

M of a rectangular square loop can also the determined by this method

cr

(c)

ac ki

itj

anticlockwise becomes north pole [N].

M  iA



= i AB  BC



  = i CD  DA  = i BC  CD



= i DA  AB



Example. A square loop of side

carries a current i. Find the magnetic moment

of the loop. For more Study Material and Question Bank visit www.crackiitjee.in

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Solution: Method-I:

M  iA n where 2

Mi and

3i j  2 2

M 

i

2

2





3 i j

itj

=

ac ki

Method-II:





cr

M  i OC  CB Here

ee .in

n  sin60 i  sin60 j

OC  cos 60 i  sin60 j =

i 3 j  2 2

=

i 3  j   2 2  

and

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M  i

=i

=

i

2

2

 i 3   2  2

  j  k   

 j 3   i  2 2  

2

 3 i  j Ans.  2 

Example. Find the magnetic moment of the current carrying loop ABCDEFA. Each side of the loop is

and current in the loop is i.

ee .in

Solution:

The given loop can break into two loops to get magnetic moment that is

itj

in ABEFA and BCDEB.

M = iA =i

2

cr ac ki

The magnetic moment of each loop

Their magnetic moments are perpendicular to each other. Hence Mnet =

2 M 2 i

2

Ans.

Example. A thing uniform ring of radius R carrying uniform charge q and mass m rotates about its axis with angular velocity . Find the ratio of its magnetic moment and angular momentum. Solution: The equivalent current in the ring

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i=

q q q   T 2 2 

Magnetic moment M = iA

q  R 2 2

=

L = I





= mR2 

M q  Ans. L 2m

cr ac ki



itj

Angular momentum

ee .in

qR 2 = 2

Example. A positive charge q is distributed over a circular ring of radius a. It is placed in a horizontal plane and is rotated about its axis at a uniform angular speed . A horizontal magnetic field B exists in the space. Find the torque acting on the ring due to the magnetic force. Solution: We know that  = MB sin  Here

  90

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   MB  i AB

   q  =  a2 B  2    =

1 qa2 B Ans. 2

Example. A thing insulated wire forms a plane spiral of N light turns carrying a current i. The radii of inside and outside turns are equal to a and b. Find

.in

the magnetic moment of the spiral with a given current.

ee

Solution:

Consider dr width of the spiral, at a distance r from the centre. Number

N dr b  a

ac ki

dN =

itj

of turns in this

cr

The magnetic moment of current loop, having dN turns is, dM = (dN) iA

Total magnetic moment M=

 dM

b

=

 dN iA a

 N  =  b  a

b

 dr i  r

2

a

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 Ni = b  a =

b

r

2

dr

a

 Ni b3  a3 Ans. 3 b  a





Example . A circular coil with 250 turns, an area A of 2.52 × 10 –4 m2, and a current of 100 µ A. The coil is at rest in a uniform magnetic field of magnitude B = 0.85 T with its magnetic dipole moment M initially aligned with B. What is the direction of the current in the coil?

(b)

How much work would be torque applied by an external agent

ee .in

(a)

to do on the coil to rotate is 90° from its initial orientation, so

itj

that M is perpendicular to B and the coil is again at rest?

(a)

cr ac ki

Solution:

According to the right-hand rule, the direction of the current through the wires from the right side of the coil is from top of bottom (see figure).

(b)

Work done by the agent is given by 2

W=

  d

1 2

=

 MB sin  d

1

= MB  cos 1  cos 2 

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1  0, 2  90  W  MB cos 0  cos 90 = MB = NiAB = (250) (100 × 10–6)(2.52 × 10–4) (0.85) = 5.4 × 10–6 J

cr

ac

ki itj e

e. in

= 5.4 µ J Ans.

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Galvanometer : Ammeter and Voltmeter 1.

Moving-coil Galvanometer It is an easy and sensitive instrument to detect and measure electric current. Its action is based upon the principle that when electric current flows in a coil placed in a magnetic field, a deflecting torque acts upon the coil whose magnitude depends upon the strength of the current. From the measurement of the deflection of the coil, the strength of the current can be computed.

ee .in

Moving-coil galvanometers are of two types: suspended-coil galvanometer and pivoted-coil (or Weston) galvanometer. The principle and working of the two types of galvanometers are the same, only there is some difference in their constructions. IN suspended-coil

itj

galvanometer the coil is suspended by a thin phosphor-bronze strip and

cr ac ki

the deflection of the coil is measured by a lamp and scale arrangement. In pivoted-coil galvanometer the coil swings between two pivots and the deflection of the coil is read by a pointer on a circular scale. The pivotedcoil galvanometer is less sensitive than the suspended-coil galvanometer, but can be taken easily from one place to another. (i)

Suspended-coil Galvanometer: Its construction is shown in Fig. 1. It consists of a rectangular, or circular, coil made by winding a large number of turns of fine insulated copper wire on a light, non-magnetic, metallic (such as aluminium) frame. The coil is suspended by a thin phosphor-bronze strip from a torsion-head which is connected to a terminal screw T1. To the lower portion of the strip is attached a small light mirror which rotates along with the strip and its deflection can be measured by a lamp and scale arrangement. For more Study Material and Question Bank visit www.crackiitjee.in

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The lower end of the coil is connected to a fine, loosely-wound, spring which is connected to another terminal screw T2. The coil hangs in the space between the pole-pieces of a powerful permanent horse-shoe magnet NS. The pole-pieces of the magnet are cut concave cylindrical. In between the pole-pieces, within the coil, is fixed a soft-iron cylindrical piece which is called the 'core'. The core does not touch the coil anywhere. The whole arrangement is enclosed in a non-magnetic box to protect it from air draughts. The front side of the box is made of glass. Three leveling-screws are provided at the base of

.in

the box.

ee

Theory:

When the terminal screws T1 and T 2 of the galvanometer are connected

itj

to a current-carrying circuit, a current flows in the coil of the

ac ki

galvanometer. Since the coil is suspended in a magnetic field, a deflecting torque acts upon it. Due to this torque the coil starts rotating from its position. Suppose, at any instant, the normal to the plane of the

cr

coil makes an angle  with the direction of the magnetic field. If N be the number of turns in the coil, A the area of the plane of the coil and B the magnitude of the magnetic field, then the instantaneous deflecting torque is given by

  Ni AB sin , where i is the current in the coil. As the coil rotates, he suspension strip and the spring are twisted and an elastic restoring torque is developed in them and it opposes the rotation of the coil. This torque is proportional to the angle of rotation of the coil. hence the restoring torque becomes equal to the deflecting torque, the coil rests in the equilibrium position. For more Study Material and Question Bank visit www.crackiitjee.in

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Suppose, in the equilibrium position, the angle of twist in the suspension strip = (= angle of deflection of the coil) is  radian. The deflecting torque acting on the coil will be NiAB sin . If the restoring torque for an angle of twist of 1 radian be c, then for  radian it will be c. In the equilibrium position of the coil, we have

c NAB sin 

or i 

 . sin 

Radial Field :

ee

or i 

.in

NiAB sin   c

itj

According to the above formula, the strength of the current is not

ac ki

proportional to the deflection . Hence the current cannot b read directly. to remove this difficulty, it is necessary that in every position of the coil the plane of the coil be parallel to the magnetic field so that the

cr

deflecting torque always be NiAB. It is for this purpose that the coil is suspended between cylindrically-cut pole-pieces and a soft-iron core is placed within the coil without touching it anywhere. Then the magnetic lines of forces are no longer parallel but their distribution becomes. Due to the large permeability of soft-iron, more lines of force pass through the core and thus makes the magnetic field strong. Such a magnetic field is called 'radial field'. In this field, in every position, the plane o the coil remains parallel to the liens of force i.e. in every position  = 90°. Hence the deflecting torque on the coil suspended in radial magnetic field is always NiAB. Then, in equilibrium position, we have

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NiAB  c or i 

c  NAB

…(i)

or i  K.



where K   

c  is a constant which is called the 'reduction factor' of NAB 

the galvanometer. Thus,

i  ,

.in

Sensitivity:

ee

A galvanometer is said to be sensitive if a small current passed through it produces a sufficiently large deflection. The sensitivity is of two types:

itj

current sensitivity and voltage sensitivity.

ac ki

The current sensitivity of a galvanometer is defined as the deflection produced in the galvanometer when a unit current flows through it.

cr

If a current i produces a deflection  in the galvanometer, then the current sensitivity is

 NAB  . i c

 . By eq. (i), we have i …(ii)

The reverse of current sensitivity is known as 'figure of merit' of the galvanometer. The voltage sensitivity of a galvanometer is defined as the deflection produced in the galvanometer when a unit voltage is applied across its coil. For more Study Material and Question Bank visit www.crackiitjee.in

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If a voltage produces a deflection  in the galvanometer, then the voltage sensitivity is

 . If R be the resistance of the coil of the V

galvanometer and i the current through it, then V = iR. Thus, the voltage sensitivity is

   . V iR

…(iii)

cr ac ki

 NAB  . V cR

ee .in

we have

 from eq. (ii), i

itj

Putting the value of

Eq. (ii) and (iii) show that the sensitivity of a galvanometer can be increased by increasing the values of N, A and B, and decreasing the value of c.

However, N and A cannot be increased beyond a limit, otherwise the resistance and also the mass of the galvanometer coil would increase undesirably. Therefore, B is made as large as possible and c as small as possible. The magnetic field B is increased by taking a very strong horseshoe magnet and placing a soft-iron core within the coil. The core concentrates the liens of force. The torsional constant c is made small by using a long and fine suspension strip and a thin spiral spring of phosphor-bronze or quartz. The quartz fibres are silvered to make them conducting.

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A suspended-coil galvanometer can measure current of the order or 10 – 19

ampere.

In the suspended-coil galvanometer the suspension strip is very delicate and may be broken by a small jerk. Therefore, the galvanometer cannot be taken from one place to another. This difficulty has been removed in the Weston galvanometer.

Pivoted-coil (or Weston) Galvanometer:

ee .in

This is also a moving-coil galvanometer. It is less sensitive than a suspended-coil galvanometer but is more convenient. It consists of a coil having a large number of turns of fine insulated copper wire wound on an aluminium frame. The ends of the axle of the frame of the coil are

itj

inserted in two pivots so that the coil may rotate about the axle. At both ends of the coil, near the pivots, are attached two springs which produce

cr ac ki

(ii)

torsional couple on the rotation of the coil, and connect the coil to two terminal screws T1 and T2. On both sides of the coil there are pole-pieces of a permanent strong horse-shoe magnet. The coil rotates in the magnetic field of these pole-pieces. to read the deflection f the coil, a pointer is attached to the coil which moves over a circular scale. The divisions on the scale are made at equal distances and the zero-point is in the centre. That is why no positive and negative signs are marked at the terminal screws of the galvanometer. To make the magnetic field radial, the pole-pieces are cut cylindrical and a soft-iron core is placed within the coil. It principle and working in similar to that of a suspendedcoil galvanometer. It can measure currents upto 10 –6 ampere.

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2.

Shunt The galvanometer is used in electrical circuits to detect the current, and in experiments to determine the null point. If somehow heavy current happens to flow in the galvanometer then due to very large deflection the pointer may strike and 'stop pin' and be broken, or the coil of the galvanometer may burn due to excessive heat produced. The save the galvanometer from these possible damages, a thick wire or a strip of copper is connected in parallel with its coil. It is called the 'shunt'. Its resistance is very small compared to the resistance of the coil. Therefore,

ee .in

most of the part of the current goes through the shunt and only a very small part goes through the coil. Hence there are no chance of the buring of the coil or the breaking of the pointer.

itj

Shunted-galvanometer is very useful for determining the null-point in

cr ac ki

meter-bridge and potentiometer experiments. First, the approximate position of the null-point is determined by using the shunted galvanometer. At this stage the current in the circuit is very feeble. Now the shunt is removed from the galvanometer so that full current goes through the galvanometer and accurate position of the null-point is determined.

Ammeter

An ammeter is an instrument used to measure current in an electric circuit directly in ampere. The instrument measuring currents of the order of milliampere is called milli-ammeter.

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Ammeter is essentially a galvanometer which is inserted in the circuit in series so that whole of the current in the circuit passes through it .The deflection produced in the ammeter is a measure of the current. Since, however, the coil of the ammeter has some resistance, so on connecting it in series of the circuit the resistance of the circuit increases and the current in the circuit somewhat decreases. Therefore, the current read by the ammeter is less than the actual current to be measured. Hence it is necessary that the resistance of the ammeter itself should be as small as possible so that on connecting it in the circuit the

An ideal ammeter has zero resistance.

ee .in

current to the measured may not fall appreciably.

cr ac ki

itj

Conversion of Galvanometer into Ammeter

A galvanometer as such cannot be used as an ammeter because it has appreciable resistance and it can measure only a limited current corresponding to the maximum deflection on its scale.

In practice, an ammeter is made by connecting a low resistance S in parallel with a pivoted – type moving-coil galvanometer G. S is known as ‘Stunt’. Its value depends upon the range of the required ammeter and can be calculated as follows:

Let G be the resistance of the galvanometer and ig the current which, on passing through the galvanometer, produces full scale deflection. If I is the maximum For more Study Material and Question Bank visit www.crackiitjee.in

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current to the measured, then a part i g of the current should pass through the galvanometer G and the rest (i – ig) through the stunt S. Since G and S are in parallel, the potential difference across them will be the same. ig X G = (i – ig) X S

Or, That is, only

th part of the total current will flow in the coil of the

ee

.in

galvanometer. A gain from eq. (i), we have

itj

If the current ig in the coil produces a full – scale deflection, then the current i in

ac ki

the circuit corresponds to the full – scale deflection. Thus, with a stunt S of the

5. Voltmeter

cr

above value, the galvanometer will be an ammeter of range 0 to i ampere.

A Voltmeter is an instrument used to measure the potential difference between two – points in an electrical circuit directly in volt. The instrument measuring difference of the order of milli-volt is called milli-voltmeter.

Voltmeter is also essentially a galvanometer which is connected in parallel across two points in the circuit between which the potential difference is to be measured. A part of the circuit current passes through the coil of the voltmeter For more Study Material and Question Bank visit www.crackiitjee.in

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and the p.d. between its ends equals the p.d. between those points. The deflection produced in the voltmeter is proportional to the current flowing in its coil and hence to the potential difference between its ends. Thus, the deflection is a measure of the potential difference. Since, however, the coil of the voltmeter has a finite resistance and draws some current, so on connecting it across two points, the p.d. between those points somewhat falls. Therefore, the p.d. read by the voltmeter is slightly less than the actual value to be measured. Hence the resistance of the voltmeter should be as high as possible so that on connecting it in the circuit across two points the p.d. to be measured may not fall

e. in

appreciably.

iit

je

6. Conversion of Galvanometer into Voltmeter

ac k

A voltmeter is made by connecting a high resistance R in series with a pivoted – type moving – coil galvanometer G. The value of R depends upon the range of the

cr

required voltmeter and can be calculated as follows:

Let G be the resistance of the galvanometer and ig the current which on passing through the galvanometer, produces full-scale deflection. Suppose, V is the maximum p.d. to be measured which exists between two points a and b in the circuit. On connecting the galvanometer across a and b, a current i g flows through it. Then , by Ohm’s Law, we have

Or, G+R =

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Or, R = If the current ig in the coil produces a full scale deflection, then for the potential difference V between a and b there will be a full-scale deflection. Thus, on connecting a resistance R of the above value in series with the galvanometer, the galvanometer will become a voltmeter of range 0 to V volt.

For the voltmeter, a high resistance is connected in series with the galvanometer and so the resistance of voltmeter is very high compared to that of a

cr

ac

ki itj e

e. in

galvanometer.

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ELECTROMAGNETIC INDUCTION AND ALTERNATING CURRENTS

ELECTROMAGNETIC INDUCTION

Magnetic Flux If we consider a plane perpendicular to a uniform magnetic field, then the product of the magnitude of the field and the area of the plane is called the 'magnetic flux' () linked with that plane. A plane of area A

ee

linked with this plane is given by

.in

placed perpendicular to a uniform magnetic field B. The magnetic flux

itj

  BA.

ac ki

If the magnetic field B , instead of being perpendicular to the plane, make an angel  with the perpendicular to the plane, then the magnetic flux linked with the plane will be equal to the product of the component

cr

1.

of the magnetic field perpendicular to the plane and the area of the plane. Thus

  B cos  A  BA cos .  is positive if the outward normal to the plane is in the same direction as B. It is negative if the outward normal is opposite to B. The SI unit of magnetic flux  is ;weber' (WB). Since B =

 , the A

magnetic field is also expressed in 'weber/meter2,' (Wb-m–2). That is why the magnetic field induction B is also called the 'magnetic flux density'. For more Study Material and Question Bank visit www.crackiitjee.in

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The CGS unit of  is 'maxwell', where I weber = 108 maxwell. Magnetic flux  is a scalar quantity (while magnetic flux density is a vector quantity). Dimensions of  : From the formula F = BiL (force on a current-carrying conductor in a



FA . il

cr ac ki

= ML2 T 2A 1  .

itj

MLT 2  L2      A L 

ee .in

magnetic field), we have B = F/iL. Thus

Magnetic flux may also be expressed in terms of the magnetic lines of force. We can represent a magnetic field by magnetic lines of force. If we draw limited lines of force so that in a magnetic field B = 1 Wb-m–2 only one line of force passes per meter2 through an area perpendicular to B, in a field B = 2 Wb-m–2 only two lines of force pass per meter2 perpendicular to B, and so on, then these lines are called the lines of flux. In a magnetic field the number of lines of flux passing per meter 2 through an area perpendicular to the field is equal to the magnetic flux linked with that plane. If a plane is parallel to the magnetic field, then no flux-line will pass through it and the magnetic flux linked with that plane will be zero.

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2.

Electromagnetic Induction Faraday, in 1831, discovered that whenever the number of magnetic lines of force, or magnetic flux, passing through a circuit changes, an e.m.f. is produced in the circuit. If the circuit is closed, current flows through it. The e.m.f. and the current so produced are called 'induced e.m.f.' and 'induced current', and last only while the magnetic flux is changing. This phenomenon is known as 'electromagnetic induction.' The magnetic flux through a circuit may be changed in a number of ways; e.g. (i) by moving a magnet relative to the circuit, (ii) by changing current

.in

in a neighbouring circuit, (iii) by changing current in the same circuit, (iv)

ee

by rotating a coil in a magnetic field. Let us consider two experiments demonstrating electromagnetic induction.

itj

Expt. 1:

ac ki

A magnet and a coil connected to a galvanometer. When the magnet is quickly moved towards the coil with its north pole facing the coil, the

cr

galvanometer deflects while the magnet is moving. This indicates a momentary current in the coil. When the magnet is withdrawn, the galvanometer again deflects, but in the opposite direction which means that the current in the coil is now in the opposite direction. If the experiment is repeated with south pole of the magnet facing the coil, the deflections are reversed.

It is further observed that faster the motion of the magnet, larger are the deflections. If the magnet is kept stationary and the coil is moved towards or away from the magnet, even then there is a deflection in the galvanometer. For more Study Material and Question Bank visit www.crackiitjee.in

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This shows that the current in the coil is produced due to relative motion between the coil and the magnet; it does not matter whether the coil is moving or the magnet is moving. Expt. 2: A primary coil P connected to a battery, and a secondary coil S connected to a galvanometer. When a current in primary coil is started by closing the key K, the galvanometer deflects momentarily, showing an induced current in the secondary in a direction opposite to that in the

a momentary current in

ee .in

primary. Similarly, when the current in primary is stopped there is again

the secondary, but in the same direction as in primary. Similar effects are observed while increasing or decreasing the primary current or changing

cr ac ki

itj

the relative position of the coils. When the coils are wound

on a piece of iron, the strength of induced current is much more increased. The induced current is increased still more when the coils are wound on the same closed iron ring.

3.

Faraday's Laws of Electromagnetic Induction The results of Faraday's experiments on electromagnetic induction have been summed up in the form of two laws which are known as the laws of Faraday's electromagnetic induction: First Law:

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When the magnetic flux through a circuit is changing, an induced emf is set up in the circuit whose magnitude is equal to the negative rate of change of magnetic flux. This is also known as Neumann's law. If  be the change in magnetic flux in a tie-interval t, then the emf induced in the circuit is given by

e

 . t

In the limit t  0,

e

ee .in

we write

d . dt

itj

If d is in weber and dt in sec, then e will be in volt. This equation

cr ac ki

represents an independent experimental law which cannot be derived from other experimental laws.

If the circuit is a tightly-wound coil of turns, then the emf will be induced in each turn and the emf's of all the turns will be added up. So, the induced emf in the whole coil will be

e  N

 N    . t y

N is called the 'number of magnetic flux linkages' in the coil. Its unit is 'weber-turns'. Second Law: The direction of the induced emf, or current, is such as to oppose the change that produced it. This is also known as Lenz's law.

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Explanation of Neumann's Law: Let us consider the magnet and coil experiment. When the north pole of a magnet is near a coil, a definite number of lines of magnetic flux originating from the pole passes through the coil. If we move either the coil or the magnet, the number of flux lines, that is, the magnetic flux passing through the coil changes. On taking the magnet away from the coil, the number of flux lines passing through the coil decreases; while on bringing the magnet nearer the coil, the number of flux lines through the coil increases. In either case, an induced emf is set up in the coil during

ee .in

the motion of the magnet. Faster the motion of the magnet, greater will be the rate of change of flux and higher will be emf induced. If both the magnet and the coil are stationary or both are moving in the same

itj

direction with the same velocity, there will be no change in flux and

cr ac ki

hence no emf will be induced.

If the coil is an open circuit (infinite resistance), emf will still be induced but there will be no current. This shows that the change in flux induces emf, not the current. Explanation of Lenz's Law: In the magnet and coil experiment, the direction f the induced current is in accordance to Lenz's law, that is, it opposes the motion of the magnet which produces it. When the north pole of the magnet is moved towards the coil, the induced current flows in a direction so that the near (left) face of the coil acts as a magnetic north pole. The repulsion between the two poles opposes the motion of the magnet towards the coil. Similarly, For more Study Material and Question Bank visit www.crackiitjee.in

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when the north pole of the magnet is moved away from the coil, the direction of the induced current is such as to make the near face of the coil a south pole. The attraction between the two poles opposes the motion of the magnet away from the coil. In either case, therefore, work has to be done in moving the magnet. It is this mechanical work which appears as electrical energy in the coil. If the direction of the induced current were such as not to oppose the motion of the magnet, then we would be obtaining electrical energy continuously without doing any work, which is impossible. Thus Leng'z

ee

.in

law is in accordance with the principle of conservation of energy.

Direction of Induced Current: Fleming's Right-hand Rule:

itj

"If we stretch the right-hand thumb and two nearby fingers

ac ki

perpendicular to one another, and the first finger points in the direction of magnetic field and the thumb in the direction of motion of the

cr

conductor, then the middle finger will point in the direction of the induced current". 4.

Induced Current and Induced Charge If in a coil of N turns the rate of change of magnetic flux be

 , then t

the induced emf in the circuit is

e  N

 . t

(Faraday's law) If the coil be closed and the total resistance of its circuit be R, then the induced current in the circuit will be For more Study Material and Question Bank visit www.crackiitjee.in

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i

e N   . R R t

It is clear from this equation that the induced current in the circuit depends upon the resistance (whereas the induced emf is independent of resistance). the charge flowed through the circuit in a time-interval

t will be given by q  i  t

N   t R t

=

N  R

=

number of turns×change in magnetic flux resistance

itj

ee

.in

=

ac ki

Substituting  in weber and R in ohm, q will be in coulomb. The above equation shows that the induced charge does not depend upon the timeinterval. Whether the change in magnetic flux be rapid or slow, the

5.

cr

charge in the circuit will remain the same. Motion of a Conductor in a Magnetic Field: Induced Potential Difference Suppose a thin conducting-rod JK of length l is situated in a uniform magnetic field of magnitude B perpendicular to the plane of paper directed downward. B has been represented by crosses (×). Suppose the rod is moved in the plane of paper perpendicular to its own axis, with a constant velocity v towards right. We know that there are always some free electron (negative charge) in a conductor. The atoms from which these electrons are detached have For more Study Material and Question Bank visit www.crackiitjee.in

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excess of positive charge. Thus the conductor has some positive and some negative charges. Of these, only the negative charges can move about within the conductor. With the conducting-rod moving with velocity v, these positive and negative charges also move in the magnetic field B towards right along with the rod. We have read that when a charge q moves in a magnetic field B with a velocity V perpendicular to the field, it is acted upon by a magnetic force Fm, where Fm = qvB. The force Fm is called the 'Lorentz force' and its direction is perpendicular

ee .in

to both B and v. According to the Fleming's left-hand rule, the force Fm on the positive charge will be directed towards J, and on the negative charge towards K. Since only the electrons are free to move within the

itj

rod, they start moving towards the end K of the rod. Thus there becomes

cr ac ki

a deficiency of electrons near the end J of the rod and this end becomes positively-charged. Simultaneously there becomes an excess of electrons at the end K and this end becomes negatively-charged. Hence an electric potential difference V is induced between the ends of the rod, due to which an electric field E is created with the rod, where

E

V . l

…(i)

The direction of E is from J (positive charge) to K (negative charge). The field E imposes an electric force Fe on each charge (q) within the rod, where Fe = qE. For more Study Material and Question Bank visit www.crackiitjee.in

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The direction of Fe on the positive charge will be towards K and on the negative charge towards J. Thus the direction of electric force Fe on each charge is opposite to that of the magnetic force Fm. As more and more electrons reach the end K, the magnitude of the force Fe goes on increasing until it equal Fm. Now, the resultant force on each charge in the rod becomes zero and the movement of electrons towards K stops. In this condition, Fe = Fm

.in

or qE = qvB or E = vB.

….(ii)

ac ki

V  v Bl.

itj

ends of the rod is given by

ee

Comparing (i), and (ii), the induced potential difference between the

in volt.

cr

if v be in meter/second, B in weber/meter2 and l in meter, then V will be

It is clear from above that when a straight conductor of length l moves with a velocity v perpendicular to a magnetic field of magnitude B cutting its flux-lines, a potential difference v B l is induced between the ends of the conductor. If the direction of the velocity v of the conductor makes an angle with the direction of the magnetic field B, then the component of v perpendicular to B will be v sin . Therefore, in this case, the potential difference induced between the ends of the conductor will be v B l sin . Clearly, if the conductor moves parallel to the field ( = 0), not potential difference will be induced. For more Study Material and Question Bank visit www.crackiitjee.in

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Explanation of Electromagnetic Induction in terms of Lorentz Force : Proof of Faraday's Law It is evident from Faraday's experiments that whenever there is a change in the magnetic flux passing through a closed circuit, an electric current is induced in the circuit. It can be explained on the basis of Lorentz force.

Suppose a conducting-rod JK is being moved without friction of the arms of a U-shaped stationary conductor MNOL with a velocity v towards

.in

right. The conductor is situated in a magnetic field of magnitude B perpendicular to the plane of paper directed downward. Due to be

ee

motion of the conducting-rod, the free electrons present in the rod are acted upon by a magnetic (Lorentz) force Fm (= q v B) which takes and

itj

electrons from the end J of the rod to the end K. Since a closed circuit is

ac ki

available to the electrons, they drift along the path J  K O  N  J. Thus, an electric current is established in the circuit along J  N  O  K  J (anticlockwise)*. So long the rod JK is kept moving in the magnetic

cr

6.

field, the electric current continues to flow in the circuit. It means that an emf is induced in the moving rod which maintains the current in the circuit. Since this emf induced in the rod is due to the motion of the rod, it is also called as the 'motional emf'. It is due to the Lorentz forces acting on the free electrons in the moving rod.

*The direction of current in the circuit is taken opposite to the direction of motion of the electrons. Suppose, the induced emf in the moving rod JK is e and the induced current in the circuit is i. We know that when a current-carrying For more Study Material and Question Bank visit www.crackiitjee.in

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conductor is in a magnetic field, it is acted upon by a force in a direction given by the Fleming's left-hand rule. The force imposed by the magnetic field B on the current-carrying rod JK is given by F' = i l B, where l is the length of the rod JK. The force F' is directed towards left. Hence, to keep the rod moving with a constant velocity v towards right, a force F equal and opposite to F' will have to be applied on it: F = – F'

.in

= – i l B. Suppose, under the force F, the rod JK moves a distance x in a time-

itj

will be given by

= – i l B x.

cr

ac ki

W = F x

But

ee

interval t and comes in the position J'K'. Then the work done on the rod

x = v × t (velocity × time-interval).

 W  ilB v t. But i×  W = – Bv lq. This work provides the necessary energy for the flow of charge in the circuit. We know that the energy supplied by a cell in flowing unit charge through a circuit is called the mf of the cell. Hence the induced emf e in the rod JK (which is working as a cell in the circuit J N O K J) is given by For more Study Material and Question Bank visit www.crackiitjee.in

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e = W/q = – B v l. In the time-interval t. the area of the circuit increases from JNOKJ to J'NOK'J'. Hence, during this time-interval, the change in the magnetic flux passing through the circuit is given by  = magnetic field × change in area (JJ'K'K) = B × (l × x)

 x  Bl . t t But

Thus

cr ac ki

itj

x  v (velocity of the rod). t

ee .in

Hence the rate of change of magnetic flux is

  Bl v. t

…(ii)

Comparing eq. (i) and (ii), we get

e

 . t

In the limit t  0,

e

d . dt

This is Faraday's law of electromagnetic induction. For more Study Material and Question Bank visit www.crackiitjee.in

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The negative sign signifies Lenz's law. The direction of current (anticlockwise) in the circuit due to the induced emf e is such that the force imposed on the rod (due to the current) opposes the motion of the rod (the motion of the rod is the cause of the current.) 7.

Eddy Currents: Foucault, in 1895, discovered that when a solid mass of metal is placed in a changing magnetic field, or moved in a magnetic field causing a change in the magnetic flux lined with it, induced currents are set up throughout the volume of the metal. Thee currents are known as 'eddy currents'.

.in

The direction of circulation of these currents is such as to oppose the

ee

motion of the metal or the change in magnetic flux (Lenz's law). Applications:

(i)

Damping:

ac ki

itj

The eddy currents can be usefully employed in the following cases:

The damping effect of eddy current is used in some moving-coil

cr

galvanometers to make them 'dead-beat'. The coil of such galvanometers is wound on a light metal frame. When the coil and frame rotate in the field of the permanent magnet, the eddy currents set up in the frame oppose the motion so that the coil returns to zero quickly. The oscillations of a moving-coil ballistic galvanometer are stopped by shortcircuiting the coil, this being due to the induced current in the coil itself. A similar damping device is used in some instruments such as balances, ammeters and voltmeters. A copper plate is attached to the moving system of the instrument such that it moves between the poles of a permanent magnet when the system is oscillating. (ii)

Induction Furnace: For more Study Material and Question Bank visit www.crackiitjee.in

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In induction furnaces, the metal to be heated is placed in a rapidly changing magnetic field provided by a high-frequency alternating current. The eddy currents set up in the metal produce so much heat that the metal melts. The process is used in extracting the metal from ore. (iii)

Induction Motor: When a metallic cylinder is placed in a rotating magnetic field, eddy currents are set up in it. These currents, according to Lenz's law, try to reduce the relative motion between the cylinder and the field. The

ee .in

cylinder, therefore, begins to rotate in the direction of the field. This is the principle of an induction motor. (iv)

Electric Brakes:

itj

When a strong, stationary magnetic field is suddenly applied to a

cr ac ki

rotating drum, the eddy currents set up in the drum exert a torque which stops the motion of the drum. This principle is used in stopping electric trains.

Besides these, eddy currents are employed in speedometers of automobiles, energy meters and in 'deep heat' treatment of human body.

8.

Self Induction When a current flows through a coil, it produces a magnetic field and hence a magnetic flux which is linked with the coil. If the current through the coil is changed, the flux linked with the coil also changes. Therefore, an induced emf is set up in the coil and an induced current flows through it, besides the main current. According to Lenz's law, the induced current For more Study Material and Question Bank visit www.crackiitjee.in

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always opposes the change in the main current. When the main current is increased, the induced current flows opposite to the main current and opposes the increase in the main current. When the main current is decreased, then the induced current flows in the same direction as the main current and opposes the decrease in the main current.

he phenomenon of electromagnetic induction in which, on changing the current in a coil, an opposing induced emf is set up in that very coil

ee .in

is called self-induction. The induced emf is called 'back emf.' Thus, when the current in a coil is switched on, the self-induction opposes the growth of the current, and when the current is switched off, the self-induction opposes the decay of the current. This is why the self-

cr ac ki

itj

induction is also called as "inertia" of electricity.

Coefficient of Self-Induction:

Let us consider a coil of N turns carrying a current i. Let  be the magnetic flux linked with each turn of the coil. Then the number of fluxlinkages is N. If no magnetic materials (iron, etc.) are present near the coil, then the number of flux-linkages with the coil is proportional to the current i, that is,

N  i or N  L i where L is a constant called the 'coefficient of self-induction' or 'selfinductance' of the coil. By the above equation, we have

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L

N . i

…(i)

If i = 1, then L = N. Hence, the coefficient of self-induction of a coil is equal to the number of flux-linkages with the coil when unit current is flowing through the coil. If, on changing the current through the coil, the back emf induced in the

itj

 N  is the rate of change of flux (due to change of current in t

ac ki

where

 N    , t t

ee

eN

.in

coil be e then, by Faraday's law, we have

the coil.

But N  Li.

 Li i  L , t t

cr

e 

Where

l is the rate of change of current in the coil. The negative sign t

indicates that the induced emf e is always in such a direction that it opposes the change of current in the coil. From the above formula, we have

L 

e . l t

…(ii)

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If

i  1, t

then L = e (numerically). Hence, the coefficient of self-induction of a coil is numerically equal to the emf induced in the coil when the rate of change of current in the coil is unity. The SI unit of the coefficient of self-induction is 'henry' (H). Thus, the self-inductance of a coil is 1 henry when an induced emf of 1 volt is set up in the coil due to a current changing at the rate of 1 ampere per

ac ki

Dimensions of L:

ee

1 volt . 1 ampere second

itj

1 henry =

.in

second in the coil, that is,

 L   ML2 T 2 A 2  .

Self-Inductance of a long Solenoid

cr

9.

Let consider a long air-cored solenoid of length l and area of crosssection A, having N closely-wound turns. The number of turns per unit length is n =

N . Let is be the current through the solenoid. The l

magnitude of the magnetic field inside the solenoid, in SI unit, is given by

B  0

N i, l

where µ0 is the permeability constant. The magnetic flux through each turn is

  BA For more Study Material and Question Bank visit www.crackiitjee.in

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 N

= 0   iA.  l  total magnetic flux (flux linkages) through the solenoid is

 N N  N0   iA  l =

0 N2 iA l

The self-inductance of the solenoid is therefore

N i

ee .in

L=

itj

0 N2 A = . l

cr ac ki

Since L is in henry, it is clear from the equation that we may take the unit for µ0 as henry/meter (Hm –1) also.

The last expression may also be written as L = µ0 n2 lA.



N  nl

If the solenoid be wound on a core of constant permeability µ, then the

 N2 iA total magnetic flux through the solenoid is and the coefficient l of self -induction becomes.

 N2A , L= l where µ = µ0 µr, (µ r being the relative permeability). For more Study Material and Question Bank visit www.crackiitjee.in

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The corresponding expression for a plane coil of radius a will be

  N2 a L= L  . 2 Mutual Induction If we place two coils near each other and pass electric current in one of them, or change the current already passing passing through it, or stop the current, then an emf is induced in the second coil. This phenomenon of electromagnetic induction is called 'mutual induction'. The first coil is

ee .in

called the 'primary coil' and the second is called the 'secondary coil'.

A primary coil P an a secondary coil S are placed near each other. The coil P is connected to a battery, a taping key and a rheostat. The

itj

secondary coil S is connected to a galvanometer. As the key is pressed, a 'momentary' deflection is produced in the galvanometer. The reason is

cr ac ki

11.

that when the key is pressed, a current passes in the coil P and a magnetic field is produced around it. Some lines of the magnetic flux pas through the coil S also. Thus, on pressing the key, the number of fluxlines passing through the coil S increases from zero to a definite value. Due to this change in the number of flux-lines, that is, in the magnetic flux, an emf is induced in the coil S and a current flows through it. Hence a 'momentary' deflection is produced in the galvanometer. According to Lenz's law, the induced electric current opposes the creation of fluxlines. Hence its direction is opposite to the direction of current flowing in the coil P. When the key is left, again a 'momentary' deflection is produced in the galvanometer but now in the opposite direction. The reason is that on For more Study Material and Question Bank visit www.crackiitjee.in

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leaving the key, the current flowing in P stops and the flux-lines passing through S disappear, that is, the number of flux lines becomes zero. Hence again an emf is induced in the coil S and a current flows. According to Lenz's law, now this current opposes the disappearance of flux lines, that is, it tends to maintain the flux-lines. So, now the direction of current in S is same as was in P. Similarly, when the current through the coil P is varied with the help of the rheostat, there is a change in the number of flux-lines passing through S. Hence so long as the current in P is changing, an induced

.in

current in S flows.

ee

If we place cores of ferro-magnetic material (iron) in the coils or wind both the coils on the same core, then the emf induced in the secondary

ac ki

this principle.

itj

soil S is increased. The induction coil and the transformer are based on

cr

Coefficient of Mutual Induction :

Let a current of i1 ampere flow in the primary coil. Let, due to this current, the magnetic flux linked with each turn of the secondary coil be 2. If N2 be the number of turns in the secondary coil, then the number of flux-linkages in the coil will be N22. This number is proportional to the current i1 flowing in the primary coil, that is,

N2 2  i1 or N2 2  Mi1, where M is a constant called the 'coefficient of mutual induction' of the two coils. From the above equation, we have For more Study Material and Question Bank visit www.crackiitjee.in

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M

N2 2 . i1

If i1 = 1, then M = N2 F2. Hence, the coefficient of mutual induction of two coils is equal to the number of magnetic flux-linkages in one coil when a unit current flows in the 'other'. If, on changing the current in the primary coil, the emf induced in the secondary coil be e2, then according to Faraday's law,

2 t

ee

e2   N2

.in

we have

 N22  , t

where

2 is the rate of change of flux in the secondary coil (due to t

ac ki

itj

= 

cr

change of current in the primary coil). But N2 2 = Mi1.

 e2  

 Mi1  t

= M

i1 , t

where

i1 is the rate of change of current in the primary coil. The t

negative sign indicates that the direction of emf induced in the secondary coil is always such that it opposes any change in current in the primary coil. From the above expression, For more Study Material and Question Bank visit www.crackiitjee.in

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we have

M 

if

e2 . i1 t

i1  1, then M = e2 (numerically). t

Hence, the coefficient of mutual induction of two coils is equal to the numerical value of the induced emf in one coil when the rate of change

.in

of current in the other coil is unity. The SI unit of the coefficient of mutual induction is 'henry' (H). Thus, the

ee

mutual inductance of two coils is 1 henry when an induced emf of 1

itj

volt is set up in one of them due to a current changing at the rate of 1

ac ki

ampere per second in the other. 1 H = 1 Wb A–1

12.

cr

= 1 Vs A–1.

Mutual Inductance of two Coaxial Solenoids In general, the mutual inductance of two coils depends on the geometry of the coils (size, shape, number of turns, etc.), the distance between the coils, the relative orientation of the coils and the nature of the material on which the coils are wound. Let us consider a long air-cored solenoid P (primary) of length l and area of cross-section A, having N1 turns. Let a shorter secondary coil S having N2 turns be would closely over the central portion of the primary P. Let i 1 For more Study Material and Question Bank visit www.crackiitjee.in

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be the current in the primary. The magnitude of the magnetic field inside the primary is, in SI unit, given by: B = 0

N1 i. l 1

The magnetic flux linked with the primary due to the field B is also linked with the secondary because the secondary is would closely over the central portion of the primary. Thus, the magnetic flux linked with each turn of the secondary is

= 0

N1 i A, l 1

ee .in

2 = B A

where A is the area of cross-section of the secondary (although it is

itj

almost the same as of the primary).

cr ac ki

The total magnetic flux (flux linkages) through the secondary is N2 2 = N2 B A =

0 N1 N2 i1 A . l

By definition, the mutual inductance of the two solenoids is M=

N22 i1

or M =

0 N1 N2 A l

If n1 be the number of turns per unit length in the primary, then n1 =

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and we have

M  0 n1 N2 A. If an iron core be placed inside the primary, the magnitude of M would greatly increase. Coefficient of Coupling: The coefficient of coupling K of two coils is a measure of the coupling of the coils and is given by

.in

M . L1 L 2

ee

K=

where L1 and L2 are coefficients of self-induction and M the coefficient of

Inductances in Series and in Parallel

cr

13.

ac ki

K is always less than 1.

itj

mutual induction of the two coils.

Inductances in Series:

When two coils of inductances L1 and L2 are connected in series and a current i is passed through them, the total flux linkage  is the sum of the flux linkages L1 i and L2 i. That is

  L1 i  L2 i. If L be the equivalent inductance of the system, then

  L i. For more Study Material and Question Bank visit www.crackiitjee.in

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 Li  L1 i  L2 i or L = L1 + L2. This is the required expression which holds provided the two coils are separated by a large distance so that there is no flux linkage in any coil due to the current in the other. When the separation between the coils is small, there will be a mutual inductance M between them. In this case, the resultant induced emf e in the two coils is the sum of the emf's e1 and e2 in the respective coils.

.in

That is e = e1 + e2

i i   i i   M    L 2 M , t t   t t 

itj

ee

 

=  L1

ac ki

is being assumed that the coils are so placed that the flux linking each coil due to the current in itself is in the same direction as the flux due to the current in the other coil.

i . t

cr

But e  L

 L

i i i i  L1  L2  2M t t t t

or L = L1 + L2 – 2M. If, however, the coils are so placed that the flux linking each coil due to its own current is opposite in direction to the flux due to the current in the other, then L = L1 + L2 – 2M. (ii)

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Suppose two coils L1 and L2 are connected in parallel between two points, and a current i is divided between them. i = i1 + i2 or

i i1 i2   . t t t

When the currents through the inductances were growing, induced emf's were set up in them. As the potential difference across each coil is

i1 t

e =  L1

i2 . t

= – L2

ee .in

the same, the induced emf's must also have the same value e (say). Thus

or

i t

cr ac ki

e = L

itj

If L be the equivalent inductance, then

e i  L t

 i1 i2   t t 

=    =

or

e e  L1 L 2 1 1 1   L L1 L 2

or L =

L1 L 2 . L1  L 2 For more Study Material and Question Bank visit www.crackiitjee.in

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This is the required expression.

TRANSIENT CURRENTS 1.

Growth and Decay of Current in an Inductance-Resistance (L-R) Circuit

Growth of Current: Let us consider a circuit containing a resistance less coil of selfinductance L and a non-inductive resistance R connected to a battery of constant emf E through a two-way switch S. When the circuit is closed by

ee .in

throwing the switch S to a, a self-induced emf is set up in the coil which opposes the growth of current in the circuit. Hence the current does not reach its final steady value E/R instantly, but grows at a rate depending

cr ac ki

itj

upon the values of L and R in the circuit.

During the variable state, when the current is growing, let i be the current and

di the rate of growth of current at any instant t. Then, the dt

instantaneous p.d. across the) resistance R is

VR  i R and that across the inductance L is

VL   L

di . dt

Since VL is opposite to E, the net p.d. that appears across the resistance

 di  . This, by Ohm's law, must be equal to i R. Hence dt 

is E – L  

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EL

di  iR dt

or E  IR  L or dt 

di dt

L di E  iR

Integrating, we get

L loge E  iR   C, R

t

C

L loge E. R

itj

Now, at t = 0, i = 0; thus

ee .in

where C is the integration constant.

cr ac ki

Hence the last expression becomes

L L loge E  iR   loge E R R

t or 

R t  loge E  iR   loge E L

= loge

or e

E  iR E

R   t L

=1



E  iR E

iR E R

  t iR  1  e L or E For more Study Material and Question Bank visit www.crackiitjee.in

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E or i  R But

R    t  L 1  e .  

E  i0 , the final steady value of current R

R    t   i  i0 1  e  L    

...(i)

This is the Helmholtz equation for the growth of current in an L – R







E  asymptotically at R

ee

reaching its steady value i0   

.in

circuit. It shows that the current in an L – R circuit rises exponentially;

itj

t   e   0 . A graph current and time is shown in Fig. 2.

ac ki

The rate of growth of the current is obtained by differentiating eq. (i), this is,

R

cr

di R   t  i0 e  L  dt L = i0

R i 1  L  i0 

[from eq. (i)] =

R i  i . L 0

Thus it is seen that greater the ratio

R L , or smaller the ratio , the L R

more rapidly does the current approach its maximum value. The ratio For more Study Material and Question Bank visit www.crackiitjee.in

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L is called the 'inductive time-constant' L of the circuit and is R expressed in second if L and R are given in henry and ohm respectively. Putting t = L =

L in eq. (i), R

we get



= i0

e 1 e

= i0

2.718  1 2.718 e  2.718

itj



ee .in



i = i0 1  e1

cr ac ki

= 0.632 i0.

Thus, the inductive time-constant of an L – R circuit is the time in which the current grows from zero to 0.632 of its steady value. We can also see how the p.d. across the resistance and that across the inductance vary with time during the growth of current. We have VR = iR R    t  = i0 R 1  e  L    

R    t  = E 1  e  L    

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and VL  L

= i0 Re = Ee

di dt

R   t L

R   t L

Thus, VR varies with t in the same manner as i does, whereas VL varies. It is maximum (= E) at t = 0 and falls off exponentially to zero as the current

.in

approaches its maximum steady value.

ee

Decay of Current:

When the switch S is thrown over to b, the battery is cut off, the L – R

itj

circuit is again closed and the current in the circuit decays. The self-

ac ki

induced emf in the coil now opposes the decay of the current. Since, now E = 0, the equation for decay is

di  iR dt

cr

L

or dt  

L di . R i

Integrating, we get

t

L loge i  C, R

where C is the integration constant. Now, at t = 0, i = i0 ; thus

C

L loge i0. R For more Study Material and Question Bank visit www.crackiitjee.in

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Hence the last expression becomes

L L loge i  loge i0 R R

t

R t  loge i  loge i0 L

= loge

or e

i i0

R   t L



R   t L

…(ii)

itj

or i  i0e

i i0

ee .in

or 

This is the Helmholtz equation for the decay of current in an L – R circuit.

cr ac ki

It shows that the current in an L – R circuit decays exponentially, reaching zero at t (asymptotically).

The rate of decay of current is obtained by differentiating eq. (ii), that is, R

di R   t  i0 e  L  dt L = i0

R i L i0

[from eq. (ii)] = 

R i. L

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Thus, it is clear that greater the ratio constant

R , or smaller the inductive timeL

L , the more rapidly does the current decay. Show that the R

growth and decay curves are complementary. Putting t = L =

L R

in eq. (ii), we get

i0 2.718

= 0.368 i0.

ee

=

itj

i0 e

ac ki

=

.in

i  i0 e1

cr

Thus, the inductive time-constant of an L – R circuit may also be defined as the time in which the current decay from maximum to 0.368 of the maximum value. Energy of the Decay Current : When the current in a coil is switched on, the induced emf opposes the growth of current. Hence the agent (battery) supplying the current does work against the induced emf. When the current becomes stady, there is no induced emf and no more work is done The total work done in bringing the current to its final value is stored in the magnetic field of the coil. It is liberated back in the form of heat and light when the current decays. For more Study Material and Question Bank visit www.crackiitjee.in

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Self Induced E.M.F. at Break: The current in a circuit is usually cut off by breaking the circuit. This introduces into the circuit the very high resistance of the air gap. Hence the time constant,

L , is much less at break than at make. Therefore, the R

rate of decay of current at break is much greater than the rate of growth of current at make. Hence the rate of change of magnetic flux linked with the circuit is much greater at break than at make. Therefore, the induced e.m.f., which is directly proportional to the rate of change of magnetic

.in

flux, is much greater at break than at make. This emf at break is often great enough to break down the insulation of air between the switch

ee

contacts, and produces a spark. This explains the sparking across the

itj

make and break contact in an induction coil in the absence of the

Energy Stored in an Inductor

Whenever a current in an inductor is switched on, a back emf is induced in the inductor which oppose the growth of current. Therefore, the

cr

3.

ac ki

capacitors.

growing current has to work against the back emf before it attains its final steady value. This work is provided by the battery supplying the current and is stored as magnets energy around the inductor. Let i be the current in the inductor L at any instant t after it starts to grow. The instantaneous self-induced emf across the inductor is

e  L

di dt

The work required to move a charge dq against this emf is dW = – e dq For more Study Material and Question Bank visit www.crackiitjee.in

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=L

di dq dt

=L

dq di. dt

But

dq  i, the instantaneous current. dt

dW = L i di. Hence the total work required to build up the current from zero to a steady maximum value i0 is

i0

ee

=L

.in

 dW

W=

 i di

1 2

L i20.

ac ki

=

itj

0

This work is used to establish the magnetic field around the inductor

Thus

cr

where it is "stored" as potential energy U.

U  12 Li20 When the battery supplying the current is cut off, the magnetic field collapses and the energy is returned to the circuit. If the current-supplying source is an alternating source of emf, then during one-half cycle energy is stored in the inductor from the source and during the next half-cycle the same energy is returned to the source. Thus the average power of the inductor remains zero. For more Study Material and Question Bank visit www.crackiitjee.in

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We have already read that when the plates of a capacitor are connected to the terminals of a d.c. battery, the capacitor is charged. The energy stored in the charged capacitor is given by

U

1 2

2

CV 

1 2

q2  C

1 2

qV,

where q is the charge on the capacitor and V the p.d. across its plates. This (electrical) energy resides in the electric field between the plates of the capacitor.

remains zero. L–C Oscillations

itj

When a charged capacitor is allowed to discharge through a resistance less inductor, electrical oscillations of constant amplitude are produced

cr ac ki

4.

ee .in

In case of alternating source of emf, the average power of the capacitor

in the circuit. These are called L–C oscillations. Their production can be explained in the following way :

Initially, suppose, the capacitor C is fully charged to q0. The current in the inductor L is zero. At this instant an amount of energy

1 2

q20 is stored in C

the electric field between the plates of the capacitor. The energy stored in the magnetic field of the inductor is zero because the current is zero. When the circuit is closed by plugging in the key K, the capacitor begins to discharge through the inductor, causing a current to flow in the anticlockwise direction. As the current rises from zero, it builds up a magnetic field around the inductor, and when it has reached its maximum value i0, the energy stored in the magnetic field is

1 2

Li20 . At

the same time, the capacitor has completely discharged and the p.d. For more Study Material and Question Bank visit www.crackiitjee.in

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between its plates has reduced to zero. Thus, the energy of the electric field between the capacitor plates has transferred to the magnetic field. The current, however, does not stop instantaneously but continues to flow while the magnetic field dies down. By Lenz's law, the dying magnetic field induces an emf in the inductance in the same direction as the current. The current therefore persists, and now charges the capacitor in the opposite sense until the magnetic field has disappeared.

capacitor plates.

1 q20 in the electric field between the 2 C

ee .in

The energy is re-stored as

The capacitor begins to discharge again, the current now being clockwise

itj

The energy is once again transferred to the magnetic field around the

cr ac ki

inductor.

Reasoning as before, the circuit eventually returns to its initial situation The process then repeats at a definite frequency. In the ideal situation of zero resistance, the L–C oscillations continue indefinitely, energy being shunted back and forth between the electric field of the capacitor and the magnetic field of the inductor. Effect of Resistance : In practice, the inductance has always a finite resistance which causes gradual dissipation of energy as heat. Owing to this damping effect, the oscillations gradually diminish in amplitude and ultimately die away. Analogy with Linear Harmonic Oscillator : The LC oscillations are similar to the oscillations of a mass-spring system. In each case, the energy alternates between two forms; magnetic and For more Study Material and Question Bank visit www.crackiitjee.in

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electric for the L–C system, and kinetic and potential for the mass-spring system. The current in the L–C circuit is zero when the entire energy is stored as electric energy in the capacitor, and is maximum when the entire energy is stored as magnetic in the inductor. In the same way, the velocity of mass is zero when the energy is entirely potential energy in the spring, and is maximum when the energy is entirely kinetic energy of

 

the mass. Thus, the capacitor is like a spring  C 

1 and the inductor k 

is like a mass (L m). The charge corresponds to the displacement (q x)

1 q2 1 2  kx . 2 C 2

cr ac ki

and

itj

1 2 1 Li  mv2 2 2

ee .in

and the current corresponds to the velocity (i v). Further

Mathematical Treatment of L–C Oscillations : Let a capacitor of capacitance C, initially charged to q0, be discharged through an ideal inductor L by throwing the switch S over to b (Fig. 10). Suppose at an instant t during the discharge, q is the charge left on the capacitor and i is the discharging current through the inductor. The instantaneous p.d. across the capacitor is

q and that induced across the C

q di  di  . Thus the net p.d. in the circuit is  L . But  dt  C dt

inductor is L  

this must be zero because there is no resistance in the circuit. Thus,

q di L  0. C dt For more Study Material and Question Bank visit www.crackiitjee.in

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Now i  

dq dt

(minus sign enters because q is decreasing with increasing t).



q d  dq  L   0   C dt  dt

d2q q or L 2  0 dt C d2q q or  0 2 dt LC d2q  2q  0, 2 dt

1 . This is a linear differential equation df second order. LC

itj

where 2 

ee .in

or

cr ac ki

The general solution of this equation is of the following form : q = A cos t + B sin t

…(i)

where A and if are constants. There values may be found by applying initial conditions. At t = 0, q = q0. Then, from eq. (i), we have q0 = A cos 0 + B sin 0 or A = q0. Differentiating eq. (i), we get For more Study Material and Question Bank visit www.crackiitjee.in

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dq   A sin t  B cos t dt Again, at t = 0, q = q0 (maximum) and so

dq = 0. dt

 0 = – A sin 0 + B  cos 0 or B = 0. Putting A = q0 and B = 0 in eq. (i),

q  q0 cos t.

ee .in

we get

itj

This is the equation of the discharge of the capacitor. It shows that the

cr ac ki

discharge of the capacitor is oscillatory and simple harmonic. The period of oscillation is given by

T

2 

= 2 LC.

and the frequency of discharge (oscillation) is

f 

=

1 T

1 . 2 LC

Energy Conservation in L–C Oscillations:

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In an L–C circuit, during oscillations, the energy is partly electric and partly magnetic. The sum of the two is always constant (independent of time). Let q be the charge on the capacitor and i the current in the inductor at any time t. Then, the instantaneous electric energy stored in the capacitor is

1 q2 UE  . 2 C

1 . LC

where  

1 2 q0 cos2 t. 2C

itj

 UE 

ee .in

But q = q0 cos t (equation of discharge),

UM 

cr ac ki

The instantaneous magnetic energy stored in the inductor is

1 2 Li 2

1  dq = L  2  dt 

2

1 d  = L   q0 cos t   2  dt  =

1 2 2 Lq0  sin2 t 2

=

1 2 q0 sin2 t. 2C

2

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  

2 

1 LC 

The total instantaneous energy in the L–C circuit is

UE  UM 

1 2 q0 cos2 t  sin2 t 2C





1 q20 = , 2 C

cr

ac

ki itj e

e. in

which is independent of time.

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NUMERICAL EXAMPLES Quest. A coil having an inductance of 2.0 H and resistance of 10  is suddenly connected to a resistance less battery of 100 volt. Calculate (i) timeconstant of the circuit, (ii) the equilibrium current in the circuit and (iii) rate of growth of current when the circuit is just made. Solution: The inductive time-constant is

=

.in

L R

L 

2.0 10

ee

(i)

The equilibrium (steady) current is

i0 

100 V 10 

cr

=

E R

ac ki

(ii)

itj

= 0.2 s.

= 10 A. (iii)

The differential equation for L–R circuit is

EL

or

di  iR dt

di E  iR  . dt L

At the instant the circuit is just made, i = 0.

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

=

di E  dt L

100 V 2.0 H

= 50 A s–1. Quest. Show that the time for attaining half the value of the final steady

L . R 

current in an L–R series circuit is 0.6931  

.in

Solution.

ac ki

R    t  i  i0 1  e  L   ,  

itj

given by

ee

The instantaneous current during its growth in an L–R series circuit is

where i0 is the final steady current.

i 1  , i0 2

cr

For

we have R

  t 1  1  e L 2

or e

or e

R   t L

R   t L

 1

1 1  2 2

2

R t  loge 2 L 

or  

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= 0.6931

L . R 

or t = 0.6931  

Quest. A coil having a resistance of 15 ohm and an inductance of 10 henry is connected to a 90-volt d.c. supply. Find the value of the current after 0.67 sec. How long will it take for the current to attain 50% of its final value ? (e = 2.718) Solution.

ee

R    t  i  i0 1  e  L   ,  

.in

The equation for the current growing in an L–R circuit is

itj

where i0 is the final (maximum) value of the current.

=

E R

90 15

cr

i0 =

ac ki

Here

= 6.0 amp and

R 15  L 10

= 1.5 sec–1. Thus i = 6.0 [1 – e–1.5t]. At t = 0.67 sec, For more Study Material and Question Bank visit www.crackiitjee.in

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we have i = 6.0 [1 – e– 1.5 × 0.67] = 6.0 [1 – e–1]

 e  1 e 

= 6.0  

= 6.0 

1.718 2.718

= 3.79 amp.

= 0.462 sec.

ee

10 15

ac ki

= 0.6931 

L R

itj

t = 0.6931 

.in

The time taken by the current to attain 50% of its final value is

Quest. The current in a coil of self-inductance 2.0 henry is increasing according

cr

to i = 2sin t2 ampere. Find the amount of energy spent during the period when the current changes from zero to 2 ampere. Solution. The energy stored in the coil when the current rises from 0 to 2 amp is

U

=

1 2 Li 2 0

1 2 2.0 henry  2 amp   2

= 4 joule. For more Study Material and Question Bank visit www.crackiitjee.in

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Quest. A solenoid of resistance 50  and inductance 80 henry is connected to a 200-volt battery. How long will it take for the current to reach 50% of its final equilibrium value ? Calculate the maximum energy stored. Solution. The instantaneous current during its growth in an L – R circuit is given by R    t  i  i0 1  e  L   ,  

1 i, 2 0

we have R

or e

or e

R   t L

R   t L

or t 

=

cr ac ki

  t 1  1  e L 2

itj

For i 

ee .in

E  is the final equilibrium current. R



where i0   



1 2

2

L loge 2 R

80  0.69 50

= 1.104 s. The maximum energy stored in the inductance is For more Study Material and Question Bank visit www.crackiitjee.in

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U=

1 2 Li 2 0

1  E = L  2 R

2

1  200  =  80    50  2

2

= 640 J. Quest. A 2.0-H inductor is placed in series with a 10- resistor and an emf of

.in

100 V is suddenly implied to the combination. At 0.1 s after start, find the inductor. (e–0.5 = 0.61)

itj

Solution.

ee

the rate at which energy is being stored in the magnetic field around

i0 

100 V 10 

cr

=

E R

ac ki

The final steady current is

= 10 A. The current at any instant t in an L–R circuit is given by R    t  i  i0 1  e  L   .  

…(i)

Here i0 = 10 A,

R L

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=

10  2.0 H

= 5 s–1 and t = 0.1 s.  i = 10 [1 – e–0.5] = 10 [1 – 0.61] = 3.9 A. The instantaneous energy in the magnetic field around an inductor L

U=

1 2 Li . 2

cr ac ki

dU di  Li . dt dt

itj

The rate of storing energy is

ee .in

carrying instantaneous current i is given by

…(ii)

Differentiating eq. (i), we get

R

di R   t  i0 e  L  dt L = i0

R i 1  L  i0 

[From eq.(i)] =

R i  i . L 0

Substituting this value in eq. (ii), For more Study Material and Question Bank visit www.crackiitjee.in

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we get

dU  iR i0  i dt = 3.9 x 10 x (10–3.9) = 238 W. Quest. A 10 µF capacitor and a 2-megaohm resistor in series are connected to a 100-V battery. Find the time after which the charge on the capacitor reaches 90% of maximum.

ee .in

Solution. The equation for the charging of a capacitor is

itj

t    CR q  q0 1  e  ,  

For

cr ac ki

where q0 is the maximum charge.

q  90%  0.9, q0

we have

0.9  1  e or e or e

or



t CR

t CR



t CR

 1  0.9  0.1



1  10 0.1

t  loge 10 CR

= 2.3026 log10 10 For more Study Material and Question Bank visit www.crackiitjee.in

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= 2.3026 or t = 2.3026 CR = 2.3026 × (10 × 10–6 F) (2 × 106 ) = 46 s. Quest. A capacitor is being charged through a resistance of 3 megaohm. If it reaches 75% of its final potential in 0.5 s, find its capacitance. Solution. The charging equation of a capacitor, in terms of potential, is

ee

.in

t    CR V  V0 1  e  ,  

itj

where V0 is the final potential to be reached.

= 0.75,

cr

V = 75% V0

ac ki

Here

t = 0.5 s

and R = 3 x 106 . We have 

0.5

310 C 0.75 = 1  e 

or e or e



6

0.5

310 C  1  0.75  0.25 6

0.5

 

3 106 C

1 4 0.25

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or



0.5  loge 4 3  106 C



= 2.3026 × log10 4 Quest. A closed coil consists of 500 turns wound on a rectangular frame of area 4.0 cm2 and has a resistance of 50 . It is kept with its plane perpendicular to a uniform magnetic field of 0.2 weber/meter2. Calculate the amount of charge flowing through the coil if it is turned over (rotated rough 180°). Will this answer depend on the speed with

Solution.

ee .in

which the coil is rotated?

As the change in magnetic flux is

itj

  2 BA.

cr ac ki

If N is the number of turns in the coil and R is its resistance, then the charge flown through the coil is given by

q

=

=

N   R

2 NBA R



 

2  500  0.2 Wb m2  4.0  104 m2



50 

= 1.6 × 10–3 Wb –1 = 1.6 × 10–3 C. The answer does not depend upon the speed of rotation of the coil. Infact, the charge flown depends upon the total change in magnetic flux, not upon the rate of change of magnetic flux. For more Study Material and Question Bank visit www.crackiitjee.in

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Quest. A copper rod of length L is revolving with angular velocity  around its one end perpendicular to a uniform magnetic field B. Prove that the emf induced across the ends of the rod is e 

1  BL2. 2

Solution. The magnetic flux linked with an area A perpendicular to a uniform magnetic field of magnitude B is given by  = BA.

.in

Suppose the copper rod of length L revolving about its one end O is completing n revolutions per unit time. Then, the rate of change of

itj

 A B t t

ee

magnetic flux linked with the revolving rod is given by

= B × ( L2) n

ac ki

= B × (area swept by the rod per unit time)

 , where  is the angular velocity of the rod. 2

cr

Now, n =



=

   B  L2  t 2

1  BL2. 2

By Faraday's law, the emf induced across the rod is given by

e 

=

 t

1  BL2. 2 For more Study Material and Question Bank visit www.crackiitjee.in

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Quest. A rectangular coil of size 5 cm × 10 cm and 100 turns is placed perpendicular to a magnetic field of 10–2 Wb m–2. If the coil is withdrawn from the field in 40 ms, calculate the induced emf. Solution. The magnetic flux through each turn of a coil of area A placed perpendicular to a magnetic field B is  = BA. When the coil is withdrawn from the field, the flux linked with it

ee .in

becomes zero. So, the change in flux is

  0   =     BA

itj

= – (10–2 Wb m–2)(50 × 10–4 m2) = – 5 × 10–5 Wb.

cr ac ki



Time-interval for the change, t = 40 ms

= 40 × 10–3 s.

By Faraday's law, the emf induced in the coil is

e  N

 t

 5  10  =  100  5

40  103

= 0.125 V.

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Quest. A coil of area 0.04 m2 having 1000 turns is suspended perpendicular to a magnetic field of 5.0 × 10 –5 Wb m–2. It is rotated through 90° in 0.2 second. Calculate the average emf induced in it. Solution. The magnetic flux passing through each turn of a coil of area A, perpendicular to a magnetic field B is given by  = BA. On rotating the coil through 90°, the magnetic flux passing through it will

.in

become zero. Thus the change in flux is

ee

 = 0 – 

= – (5.0 × 10–5) (0.04)

ac ki

= – 2.0 × 10–6 Wb.

itj

= – BA

The time-interval t is 0.2 s.

cr

Therefore, the induced emf is

e  N

 t

 2.0  10  = 1000  6

0.2

= 0.01 V. Quest. 5.5 × 10–4 magnetic flux lines are passing through a coil of resistance 10 ohm and number of turns 1000. If the number of flux lines reduces to 5 × 10–5 in 0.1 sec, find the electromotive force and the charge flowing through the coil. For more Study Material and Question Bank visit www.crackiitjee.in

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Solution. Initial magnetic flux

1  5.5  104 Wb. Final magnetic flux

2  5  105 Wb.  change in flux

= (5 × 10–5) – (5.5 × 10–4) = (5 × 10–5) – (55 × 10–5) = – 50 × 10–5 Wb.

itj

Time interval for this change,

ee .in

  2  1

cr ac ki

t = 0.1 sec.

 induced emf in the coil is

e  N

 t

 50  10  = 1000  5

0.1

= 5 V. Resistance of the coil, R = 10 . Hence induced current in the coil is

i

e R For more Study Material and Question Bank visit www.crackiitjee.in

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=

5V 10 

= 0.5 A. This current persists only during the flux-change (0.1 s). Hence the charge passed through the coil is q = i × t

= 0.05 C.

ee .in

= 0.5 A × 0.1 s

Quest. The magnetic flux through a coil perpendicular to its plane and directed into the page varies with time t (in second) according to the relation

itj

 = 6 t2 + 7f + 1 milliweber.

cr ac ki

Find the magnitude of the emf induced in the coil at t = 2 s. Also find the current and its direction in the resistance R if R = 10 . Solution.

= 6t2 + 7t + 1 mWb.

By Faraday's law, the induced emf at t = 2 s is given by

e 

d dt

= 12 t + 7 = (12 × 2) + 7 = 31 mV. The corresponding current in the resistance R = 10 is For more Study Material and Question Bank visit www.crackiitjee.in

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i=

=

e R

31 mV 10 

= 3.1 mA. The direction of the induced current will be such that the magnetic field produced by it opposes the given field (Lenz's law). It will be from left to right.

ee .in

Quest. When a wheel with metallic spokes 1.2 m long revolves in a magnetic field of 5 × 10–5 T perpendicular to the plane of the wheel, an emf of 10–2 V is induced across the rim and the axle of the wheel. Find the rate

itj

of rotation of the wheel.

cr ac ki

Solution.

Let n be the rate of rotation of the wheel. As the induced emf is given by |e| = B ×  L2 × n or n =

e B   L2

Here |e| = 10–2 V = 10–2 Wb s–1.

n 

5  10

5

102 Wb s1 Wb m2  3.14  1.2 m  1.2 m



= 44.2 rotations-s–1.

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Quest. An air-cored solenoid of length 50 cm and area of cross-section 28 cm2 has 200 turns and carries a current of 5.0 A. On switching off, the current decreases to zero within a time-interval of 1 ms. Find the average emf induced across the ends of the open switch in the circuit. µ0 = 4 × 10–7 T m A–1. Solution. The magnetic field within the current-carrying solenoid having n turns per unit length is given by

ee .in

B = µ0 n i







200



= 25 × 10–4 T.

itj

7 1   5.0 A = 4  3.14  10 T m A  50  102 m

cr ac ki

Initially, the magnetic flux linked with the solenoid of N turns is  = NBA

= 200 × (25 × 10–4 Wb m–2) × (28 × 10–4 m2) = 14 × 10–4 Wb.

On switching off the current, the flux reduces to zero.  change in flux

  0   4 =    14  10 Wb.

This change take place in time-interval t = 1 ms = 10–3 s. Therefore, the induced emf is For more Study Material and Question Bank visit www.crackiitjee.in

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 t

e

14  10 4 Wb =  103 s = 1.4 V. Quest. A small flat search coil of area 2.0 cm2 with 25 close turns placed between the poles of a strong magnet normally to the magnetic field is suddenly snatched out of the field. The total charge flown in the coil, as measured by a ballistic galvanometer connected to the coil, is 7.5 mC.

.in

The resistance of the coil and the galvanometer is 0.50 . Find the

ee

magnitude of the field of the magnet. Solution.

itj

Let B be the magnetic field and A the area of the search coil having N

ac ki

turns. The magnetic flux linked with all the N turns of the search coil placed normal to the field is

cr

= NBA.

When the coil is snatched out, the flux drops to zero. change in flux,  = 0 –  = – NBA. If the coil is in a closed circuit of resistance R, then the induced emf causes a current and a charge flows in the circuit, given by

q

=

 R

NBA . R For more Study Material and Question Bank visit www.crackiitjee.in

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B 

qR NA

 4.5  10 C  0.50  = 25  2.0  10 m  3

4

2

= 0.75 Wb m–2. Quest. An aeroplane with a wing span of 40 m is flying horizontally with a uniform speed of 360 km hr–1. At the place of flight the earth's total magnetic field is 5.4 × 10 –5 Wb m–2 and the angle of dip is 30°. Calculate

ee .in

the emf induced across the tips of the wings. Solution.

The wings of the flying aeroplane are cutting flux-lines due to the vertical

itj

component of earth's magnetic field. So, a potential difference is

cr ac ki

induced between the tips of its wings.

If the earth's total magnetic field be Be and the angle of dip , then the vertical component of earth's magnetic field is V = Be sin

= (54 × 10–5) × sin 30° = 2.7 × 10–5 Wb m–2.

We know that when a conductor of length l moves with velocity v perpendicular to a magnetic field B, then the potential difference induced across the conductor is given by e = B v l. Here l = 40 m, For more Study Material and Question Bank visit www.crackiitjee.in

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v = 360 km hr–1 =

360  1000 60  60

= 100 ms–1 and B = V = 2.7 × 10–5 Wb m–2.  e = (2.7 × 10–5) × 100 × 40 = 0.108 V.

.in

Quest. In the given figure, a metallic rod PQ of 15 cm length is resting on two

ee

parallel rails immersed in a magnetic field B of magnitude 0.50 T, perpendicular to the page directed downward. The rails, the rod and

itj

the field are in three mutually perpendicular directions. A

ac ki

galvanometer G connects the rails through a switch K. the resistance of the closed loop containing the rod PQ is 9.0 m. Answer the following: (i)

Suppose K is open and the rod moves with a speed of 12 cm

cr

s–1 in the direction shown. Find the polarity and magnitude of the induced emf.

(ii)

Is there an excess charge built up at the ends of the rod when K is open ? What when K is closed?

(iii)

With K open and the rod moving uniformly, the net force on the electrons in the rod PQ is zero, even though they do experience magnetic force due to the motion of the rod. Explain.

(iv)

Calculate the retarding force on the rod when K is closed. For more Study Material and Question Bank visit www.crackiitjee.in

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(v)

How much power should be expanded by an external agent to keep the rod moving uniformly t 12 cm s –1 when K is closed.

(vi)

How much power is dissipated as heat in the closed circuit ? What is the source of this power ? What is the induced emf in the moving rod when the field B

(vii)

is parallel to the rails ? Solution. The magnitude of the emf induced across the rod PQ moving

ee .in

(i)

perpendicular to the magnetic field is e=vBl

cr ac ki

= 9.0 × 10–3 V.

itj

= (12 × 10–2 ms–1) × 0.50 Wb m–2 × (15 × 10–2 m)

The free electrons in the rod experience Lorentz (magnetic) force q v B directed towards the end Q of the rod (Fleming's left-hand rule) and move towards Q. Hence the end P is positive and Q negative. (Here q is taken charge on an electron). (ii)

Yes, there is an excess of negative charge at the end Q and correspondingly an excess of positive charge at P. When K is closed, an anticlockwise current flows in the galvanometer circuit.

(iii)

The excess charge at the ends of the rod sets up an electric field of magnitude E (say) which exerts an electric force qE on the electrons. This force, being equal and opposite, cancells the magnetic force due to motion of the rod.

(iv)

When K is closed, the current in the circuit is For more Study Material and Question Bank visit www.crackiitjee.in

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i

e R

9.0  103 V = 9.0  103  = 1.0 A. Now, the current-carrying rod PQ is placed perpendicular to a magnetic field of magnitude B. Hence it experiences a force of magnitude given by F = i BI

= 7.5 × 10–2 N.

ee .in

= 1.0 A × 0.50 NA–1 m–1 × (15 × 10–2 m)

The force F is directed towards left (Fleming's left-hand rule).

The power that must be expanded by an external agent to keep the rod

cr ac ki

(v)

itj

Thus, it is a 'retarding' force.

moving at 12 cm s–1 towards right against the retarding force F is given by P = Fv

= (7.5 × 10–2 N) × (12 × 10–2 ms–1) = 9.0 × 10–3 W. (vi)

The power dissipated as heat is given by P = i2 R = (1.0 A)2 × (9.0 × 10–3 ) = 9.0 × 10–3 W.

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The power expanded by the external agent equals the power dissipated as heat. Thus, the source of the power dissipate is the power of the external agent. (vii)

If the magnetic field is parallel to the rails, then the moving rod will not cut the magnetic field lines and no emf will be induced.

Quest. A rectangular loop of sides 8 cm and 2 cm with a small cut is moving out of a uniform magnetic field of 0.3 T directed normal to the loop. Find the voltage developed across the cut when the velocity of the loop is 1 cm s –1 in a direction normal to the (a) longer side, (b) shorter side. For how long does the induced voltage last in each case ?

ee .in

Solution. The magnetic flux threading a plane area A perpendicular to a magnetic field B is given by

itj

 = BA.

(a)

cr ac ki

Let l be the length and b the breadth of the loop, and v its velocity. As the loop moves out of the field, the area linked with the field changes, causing a change in By Faraday's law, the induced emf is given by

e  =B

 t

A . t

Now,

A  lv t

(area swept per second). Therefore, |e| = B l v = 0.3 Wb m–2 × (8 × 10–2 m) × (10–2 ms–1) For more Study Material and Question Bank visit www.crackiitjee.in

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= 2.4 × 10–4 V = 0.24 mV. The induced voltage persists so long the flux is changing. Thus, the duration of |e| is

b 2  102 m  v 102 ms1 = 2s. Now

A = bv. t

.in

(b)

= 0.06 mV.

ac ki

= 0.6 × 10–4 V

itj

= 0.3 × (2 × 10–2) × 10–2

ee

 |e| = B b v

The duration is

cr

1 8  102  v 102 = 8 s.

Quest. A metallic square wire-loop of side 10 cm and resistance 1  is moved with a constant velocity v in a uniform magnetic field of induction B = 2 Wb m–2, as shown. The field lines are perpendicular to the plane of the loop, directed into the page. The loop is connected to a network of resistors each of 3 . The resistances of the lead wires OS and PQ are negligible. What should be the speed of the loop so as to have a steady current of 1 mA in it ? Give the direction of current in the loop. For more Study Material and Question Bank visit www.crackiitjee.in

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Solution. The mesh of the resistances QCSA is a 'balanced' Wheatstone's bridge so that the resistance CA is ineffective. Let the equivalent resistance of the bridge be R'. Then

1 1 1 1    R' 6 6 3  R '  3. Total resistance of the circuit,

ee .in

R = 3 + 1 = 4 . The emf induced in the loop is e = Bv l, and the current in the loop is

e B vl  . R R

itj

i

v

cr ac ki

Therefore, the speed of the loop is

iR . Bl

Here i = 1 mA = 10–3 A , R = 4, B = 2 Wb m–2 and l = 10 cm = 0.1 m.

103  4 v  2  0.1 = 0.02 m s–1. For more Study Material and Question Bank visit www.crackiitjee.in

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According to the Fleming's right-hand rule, the current in the vertical side of the loop within the field is upward, that is, clockwise in the loop.

Quest. In the given circuit, E = 10 V, R1 = 1.0 , R2 = 2.0 , R3 = 3.0 and L = 2.0 H. Calculate the currents i 1, i2 and i3 (i) immediately after the switch S is pressed, (ii) after some time S is pressed, (iii) immediately after S is released, (iv) after me time S is released.

(i)

ee .in

Solution. 'Immediately' after pressing the switch S, the current in the coil L, due to self-induction, will be zero, that is,

itj

i3 = 0.

cr ac ki

The current will be only in R1 and R2 which are connected in series to complete the circuit.  i1 = i2 =

E R1  R 2

=

10 V 1.0  2.0 

= 3.3 A. (ii)

After some time the switch S is pressed, the effect of self-inductance in L will disappear and current will now be established in R 3 (and L). Resistances R2 and R3 are in parallel and their equivalent resistance is

R' 

R 2R 3 R2  R3 For more Study Material and Question Bank visit www.crackiitjee.in

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=

2.0  3.0 2.0  3.0

= 1.2 .  total resistance in the circuit is R = R1 + R' = 1.0 + 1.2 = 2.2 .

Therefore, the current through R 1 is

= 4.5 A.

ee

10 V 2.2 

itj

=

E R

ac ki

i1 

.in

The total current given by the cell will pass through R1.

cr

A part i2 of this current goes through R 2 and the rest i3 (= i1 – i2 ) goes through R3. Since R2 and R3 are in parallel, the potential difference between their ends will be same:

i2 R2  i1  i2  R3.

 i2 

=

i1 R 3 R2  R3

4.5  3.0 2.0  3.0

= 2.4 A. The current through R3 (and L) is For more Study Material and Question Bank visit www.crackiitjee.in

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i3 = (i1 – i2 ) = 4.5 – 2.7 = 1.8 A. (iii)

'Immediately' after releasing the switch S, the main circuit is broken and the cell stops giving current.

 i1  0. But R3, L and R2 still form a closed circuit. So, the current i3 in R3 (and L)

i3 = i2 = 1.8 A. (iv)

ee .in

persists for some time in this circuit due to self-induction of L. Thus

After some time of releasing the switch S, the current in the closed

That is,

cr ac ki

itj

circuit also becomes zero.

i1  i2  i3  0.

Quest. Two resistances of 100  and 200 and an ideal inductance of 10 H are connected to a 3-V battery through a key K as shown. If K is closed at t = 0, find (i) initial current drawn from the battery, (ii) initial potential drop across the inductance, (iii) final current drawn from the battery, (iv) final current through 100  resistance.

Solution. (i)

'Immediately' after closing K, there is almost no current in the inductance due to self-induction, the current is only in resistances. Thus, the initial current is For more Study Material and Question Bank visit www.crackiitjee.in

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i

3V 100  200 

= 0.01 A. (ii)

The p.d. across the inductor is same as across the 100– resistance, that is, 0.01 A × 100  = 1 V.

(iii)

When the current has become steady, the opposing emf in the

ee .in

inductance is zero and it short-circuits the 100- resistance. The resistance of the circuit is now only 200  and so the current drawn

cr ac ki

3V  0.015 A. 200 

itj

from the battery is

This is the final current in the 200- resistance. (iv)

The final current through 100  is zero.

Quest. Calculate the self-inductance of an air-cored solenoid, 1 m long and 0.05 m in diameter, having 1400 turns. Solution. The self-inductance of a long solenoid of length l meter having N turns is given by

0 N2 A L= , l where A is cross-sectional area in meter2 and µ0 is permeability of free space (µ0 = 4 × 10–7 Hm–1). For more Study Material and Question Bank visit www.crackiitjee.in

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= 4.8 × 10–3 H = 4.8 mH. Question : A copper ring is held horizontally and a bar magnet is dropped : through the ring with its length along the axis of the ring. Will the acceleration of the falling magnet be equal to, greater than or lesser than that due to gravity? Answer: When the magnet approaches the coil, in accordance with Lenz's law a current will be induced in the coil which will oppose, i.e., repel the

a

mg  F F  g g m m

ee .in

approaching magnet, so

When the magnet has passed through the coil, the coil through

itj

electromagnetic induction will oppose the change, i.e., will attract the

a

cr ac ki

magnet and hence the acceleration of the magnet.

mg  F F  g g m m

Question . Three identical coils A, B and C are placed with their planes parallel to one another. Coils A and C carry equal currents(A). Coils B and C are fixed in position and coil A is moved towards B with uniform motion. Is there any induced current in B? If no, give reason; if yes, mark the direction of induced current in the diagram.

Answer: For more Study Material and Question Bank visit www.crackiitjee.in

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Due to motion of coil A towards B, the flux linked with B due to A will increase while that due to C will remain constant. So in accordance with Lenz's law a current will be induced in coil B which will oppose the change, i.e., approach of A towards it. So a current will be induced in B which will be in opposite direction to that in A (as opposite currents repel each other). Question V. (a)

A current from A to B is increasing in magnitude. What is the direction

(b)

.in

of induced current, if any in the loop. If instead of current it is an electron, what will happen?

When current in the wire AB increases, the flux linked with the loop

itj

(a)

ee

Answer:

(which is out of the page) will increase, and hence the induced current in

ac ki

the loop will be inverse, i.e., clockwise and will try to decrease the flux linked with it, i.e., will repel the conductor AB. If instead of current the moving charge is an electron moving from left to

cr

(b)

right, the flux linked with the loop (which is into the page) will first increase and then decrease as the electron passes by. So the induced current Ii in the loop will be first anticlockwise and will change direction (i.e., will become clockwise) as the electron passes by [i.e., crosses point C]. I = Constant

I = Increasing

I = Decreasing

I = Constant

I = Increasing

I = Decreasing

Ii = Zero

Ii = Inverse i.e. CW

Ii = Direct, ACW

(A)

(B)

(C)

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Problem 8. Two concentric coplanar circular loops made of wire, with resistance per unit length 10–4 m–1, have diameters 0.2 m and 2 m. A time varying potential difference (4 + 2.5t) volts is applied to the larger loop. Calculate the current in the smaller loop. Solution: If R2 is the resistance of the larger loop, current in it will be

 4  2.5t  R2

itj

=

ee .in

V2 R2

I2 

B2 =

=

cr ac ki

and so the field at its centre, if its radius is b,

0 2I2 4 b

0 2  4  2.5t  4 bR 2

So the flux of B2 linked with the smaller loop of radius a,

1  a2B2 = 10

7

 a2 

So, e1 

2  4  2.5t  bR2

d1 dt

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107  a2  2  2.5 = bR1R 2 And hence induced current in the smaller loop I1 =

e1 R1

107  a2  5 = bR1R 2 Now if  is the resistance per unit length of the loops, according to the

ee .in

given problem: R1 = 2 a and R2 = 2 b

itj

And so,

cr ac ki

a2  5 I1 = 10 b 2a   2 b  7

= 107

5a

4 b 

2

which on substituting the given data results in I1 = 10

5  0.1

7



4  1  104



2

= 1.25 A. Ans. Problem. Two infinite long straight parallel wires A and B are separated by 0.1 m distance and carry equal current in opposite directions. A square loop of wire C of side 0.1 m lies in the plane of A and B. The loop of wire C is For more Study Material and Question Bank visit www.crackiitjee.in

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kept parallel to both A and B at a distance of 0.1 m from the nearest wire. Calculate the emf induced in loop C while the current in A and B is increasing at the same rate of 10 3 A s–1. Also indicate the direction of current in loop C. Solution: The resultant field at the strip of width dx at distance x is out of the page and has the value

0  2I1 2I2     4  r1 r2 

ee .in

Bout 

itj

  1 1 Bout  107  2I    0.1  x 0.2  x      

cr ac ki

So, d  B ds



 1 1   dx 0.1  x 0.2  x      

= 107  2I  0.1  And hence,

  1 1 I    dx 0.1  x 0.2  x     0   0.1

  0.2  10

7

0.1

 0.1  x  i.e.,   0.2  10 I log   0.2  x  0 7

 4

7 i.e.,   0.2  10 I loge    3

So, e 

d dt For more Study Material and Question Bank visit www.crackiitjee.in

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= 0.2  107 

dI  0.28 dt

= 5.6  109  103 = 5.6 µV. Ans. The direction of induced current in the loop in accordance with Lenz's law will be clockwise for the situation considered. Problem 10. A rectangular frame ABCD made of a uniform metal wire has a straight

.in

connection between E and F made of the same wire AEFD is a square of side 1 m and EB = FC = 0.5 m. The entire circuit is placed in a steadily

ee

increasing uniform magnetic field directed into the plane of the paper and normal to it. The rate of change of the magnetic field is 1 T s–1. The

ac ki

itj

resistance per unit length of the wire is 1 m–1. Find the magnitude and direction of the currents in the segments AE, BE and EF. Solution:

cr

As in case of changing field, induced emf

d dt

e

=

d BS dt

=S

dB dt

So for loops a and b we have e1 = (1 × 1) × 1 =1V For more Study Material and Question Bank visit www.crackiitjee.in

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1   1  1  2

and e2 =   =

1 V 2

The directions of induced emfs e1 & e2 and currents I1 & I2 in the two loops in accordance with Lenz's law. Now by Kirchhoff's I law at junction E, I + I2 – I1 = 0

ee

and by Kirchhoff's II law in mesh a,

.in

i.e., I = I1 – I2

i.e., 4I1 – I2 = 1

…(1)

1 1 1  I2  1  I2   I1  I2   1   0 2 2 2

cr

I2 

ac ki

while in mesh b,

itj

I1 × 1 + (I1 – I2) × 1 + I1 × 1 + I1 × 1 – 1 = 0

i.e., I1  3I2 

1 2

….(2)

Solving Eqn. (1) and (2), I1 =

7 A 22

and I2 

6 A 22

So current in segment AE, I1 

7 6 A from E to A while in BE, I2 = A 22 22

from B to E and in EF, For more Study Material and Question Bank visit www.crackiitjee.in

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I = I1 – I2 =

7 6 1   22 22 22

from F to E. Ans. Problem. A current i = 3.36 (1 + 2t) x 10–2 A increases at a steady rate in a long straight wire. A small circular loop of radius 10–3 m has its plane parallel to the wire and is placed at a distance of 1 m from the wire. The

ee .in

resistance of the loop is 8.4 x 10–4 . Find the magnitude and the direction of the induced current in the loop. Solution:

0 2I 4 d

cr ac ki

B

itj

The field due to the wire at the centre of loop,

= 107 

2I 1

So the flux linked with the loop wire

  BS 2 = B  r



= 107  2I    103



2

So emf induced in the loop due to change of current |e| =

d dt For more Study Material and Question Bank visit www.crackiitjee.in

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= 2  1013

dl dt

But as here

I  2.36 1  2t   102 so

dI A  6.72  102 dt s

And hence

= 13.44 × 10–15 V

ee .in

e = 2 x 10–13 × 6.72 × 10–2

i

e R

cr ac ki

13.44  1015 = 8.4  104

itj

And so the induced current in the loop

12

= 16  10

A. Ans.

As due to increase in current in the wire the flux linked with the loop will increase, so in accordance with Lenz's law the direction of current induced in the loop will be inverse of that in wire, i.e., anticlockwise. Problem. A bar of mass m and length l moves on two frictionless parallel rails in the presence of a uniform magnetic field directed into the page. The bar is given an initial velocity v 0 to the right and released. Find the velocity of the bar as a function of time. Solution: For more Study Material and Question Bank visit www.crackiitjee.in

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The moving rod will cut B , so an emf Bvl will be induced in it due to which a current I=

=

e R

B vl …(1) R

will flow through it. The direction of I in accordance with Lenz's law (or Fleming's right hand rule) will be from B to A.

.in

Now as a current-carrying conductor in a magnetic field experiences a force I dI  B, the rod will experience a force FM = BIl opposite to its

FM = – ma

dv dt

ac ki

i.e., BIl = m

itj

ee

motion. So the equation of motion of the rod will be,

cr

dv   as F  B Il and a  M  dt  

And on substituting the value of I from Eqn. (1), the above expression becomes

dv B2l2  v dt mR i.e.,

dv B2 l2  dt v mR

And on integrating it with initial condition v = v0 at t=0 we have For more Study Material and Question Bank visit www.crackiitjee.in

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v

dv B2 l2    v mR v0

t

 dt 0

 v B2 l2 i.e., loge     t mR  v0  i.e., v  v0e with  



t 

mR Ans. B2 l2

.in

The above expression shows that the velocity of the bar decreases exponentially with time.

ee

Problem.

itj

A pair of parallel horizontal conducting rails of negligible resistance

ac ki

shorted at on end is fixed on a table. The distance between the rails is l. A conducing massless rod of resistance R can slide on the rails frictionlessly. The rod is tied to a massless string which passes over a

cr

pulley fixed to the edge of the table. A mass m, tied to the other end of the string, hangs vertically. A constant magnetic field B exists perpendicular to the table. If the system is released from rest, calculate (a) the terminal velocity achieved by the rod, and (b) the acceleration of the mass at the instant when the velocity of the rod is half the terminal velocity. Solution: If v is the velocity of the rod at any instant, the emf induced in it will be Bvl. And as the resistance of the rod is R, the induced current will be I=

e R For more Study Material and Question Bank visit www.crackiitjee.in

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=

Bvl R

Due to this induced current a force FM = BIl

B2 l2 v = ….(1) R will act on the rod which will oppose its motion.

.in

So if T is the tension in the string, equation of motion of rod and mass m will be

ee

T – FM = 0 × a

itj

and mg – T = ma

ac ki

So eliminating T between these mg – FM = ma

FM m

cr

i.e., a = g 

and on substituting the value of FM from Eqn. (1)

B2 l2 a= g v mR i.e., a = g  with  

v 

mR

Bl

2

…(2)

So the rod will achieve terminal velocity VT when a = 0, For more Study Material and Question Bank visit www.crackiitjee.in

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i.e., g 

vT 0 

i.e., v T   g 

mgR

Bl

2

….(3) Ans.

and so for

1 2

1 g, 2

from Eqn. (2), a= g

1 g Ans. 2

cr ac ki

=

11   g  2 

ee .in

vT 

itj

v

Problem.

Two parallel wires AL and BM placed at a distance l are connected by a resistor R and placed in a magnetic field B which is perpendicular to the plane containing the wires. Another wire CD now connects the two wires perpendicularly and made to slide with velocity v. Calculate the work done per sec. needed to slide the wire CD. Neglect the resistance of all the wires. Solution: When a rod of length l moves in a magnetic field with velocity v, an emf Bvl will be induced in it. Due to this induced emf, a current I=

e R For more Study Material and Question Bank visit www.crackiitjee.in

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=

Bvl R

will flow in the circuit. Due to this induced current, the wire will experience a force FM = BIl

B2 l2 v = R

.in

which will oppose its motion. So to maintain the motion of the wire CD a force F = FM must be applied in the direction of motion, and so the work

Fv = FMv

itj

dW dt

ac ki

P=

ee

done per sec, i.e., power needed to slide the wire,

Problem.

cr

B2 v2 l2 = Ans. R

An infinitesimal bar magnet of dipole moment M is pointing and moving with speed v in the x-direction. A closed circular conducting loop of radius a and negligible self-inductance lies in the y-z plane with its centre at x = 0 and its axis coinciding with x-axis. Find the force opposing the motion of the magnet, if the resistance of the loop is R. Assume that the distance x of the magnet from the centre of the hop is much greater than a. Solution: For more Study Material and Question Bank visit www.crackiitjee.in

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The field due to the magnet at the position of loop (as shown in Fig. 14.49) will be B=

0 2M 4 x3

So the flux linked with the loop,

  BS =

0 2M  a2 3 4 x

ac ki

0 6Ma2 dx = 4 x 4 dt

ee

d dt

itj

e= 

.in

And hence induced emf in the loop due to the motion of the magnet,

0 6Ma2 = v …(1) 4 x4

I=

=

e R

cr

And hence induced current in the loop,

0 6Ma2 v …(2) 4 Rx 4

So the dipole moment induced in the coil, M' = IS



= I a2



0 62a4M v …(3) = 4 Rx4 For more Study Material and Question Bank visit www.crackiitjee.in

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Now as in case of two dipoles with their axes along the line joining their centres, F=

0 6MM' 4 x4

So in the light of Eqn. (3),

9 20 M2a4 F= v Ans. 4 R x8 Problem.

ee .in

Two parallel vertical metallic rails AB and CD are separated by 1 m. They are connected at the two ends by resistances R 1 and R2. A horizontal metallic bar of mass 0.2 kg slides without friction, vertically down the rails under the action of gravity. There is a uniform horizontal

itj

magnetic field of 0.6 T perpendicular to the plane of the rails. It is

cr ac ki

observed that when the terminal velocity is attained, the power dissipated in R1 and R2 are 0.76 W and 1.2 W respectively. Find the terminal velocity of the bar and the values of RI and R2. Solution:

The rod will acquire terminal velocity only when magnetic force FM = BIl due to electromagnetic induction balances its weight, i.e., BIl = mg, i.e., I =

=

0.2  9.8 0.6  1

9.8 A 3

Now if e is the emf induced in the rod, For more Study Material and Question Bank visit www.crackiitjee.in

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e×I=P = P1 + P 2

0.76  1.20

So, e =

 9.8   3 

= 0.6 V Now as this e is generated due to motion of rod with terminal velocity in the magnetic field, i.e.,

itj

0.6 0.6  1

ac ki

=

e Bl

ee

so vT =

.in

e = BvTl

= 1 m s–1 Ans.

V2 P= R

cr

Further, as in case of Joule heating,

i.e., R 

V2 P

And as here, V1 = V2 = e

e2 So, R1 = P1

0.6  =

2

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=

9  Ans. 19

e2 And, R2 = P2

0.6  =

2

1.2

= 0.3 Ans. Question.

ee .in

What will happen to the inductance of a solenoid (a) when the number of turns and the length are doubled keeping the area of cross-section same, (b) when the air inside the solenoid is replaced by iron of relative

itj

permeability µr?

cr ac ki

Answer: In case of a solenoid as B = µ0 nI,

 = B(nlS) = µ0n2 lSI

and hence L=

 O

= 0 n2 lS

N2 = 0 S l

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N   as n  l  So (a) when N and l are doubled, L' = 0

2N 2l

2

S

N2 = 20 S  2L l i.e., inductance of the solenoid will be doubled.

L' = µn2 l S and hence,

cr ac ki

L'    r L 0

ee .in

When air is replaced by iron, µ0 will change to µ, so that

itj

(b)

i.e., L' = µr L

So inductance will become µr times of its initial value. Problem.

A small square loop of wire of side I is placed inside a large square loop of wire of side L (>> l). The loops are coplanar and their centres coincide. What is the mutual inductance of the system? Solution: Considering the larger loop to be made up of four rods each of length L,

L

the field at the centre, i.e., at a distance   from each rod, will be  2

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B= 4

0 I 4 d

sin   sin  i.e., B  4 

i.e., B1 =

0 I  2 sin 45 4  L    2

0 8 2 I 4 L

.in

So the flux linked with smaller loop

0 8 2 2 l I 4 L

itj

=

=2 2

2 l2 Ans.  L

cr

2 I

ac ki

and hence, M=

ee

2  B1S2

Problem. The current (in ampere) in an inductor is given by I = 5 + 4 + 16t, where t is in second. The self-induced emf in it is 16 mV. Find: (a)

the self-inductance, and

(b)

the energy stored in the inductor and the power supplied to it at t = 1.

Solution: (a)

Induced emf in an inductor is given by For more Study Material and Question Bank visit www.crackiitjee.in

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induced =  L

di dt

Hence

L

d 5  16t   10 mV dt

or L = 6.25 × 10–4 H Energy stored in the inductor =

1 2 Li 2

=

1 2  6.25  10 4 5  16t  2

ee .in

(b)

At t = 1 s

itj

1 × 6.25 × 10–4 × (21)2 2

cr ac ki

Energy =

= 0.1378 J

Power = VI

= 10 x 10–3 × (5 + 16t) At t = 1s

P = 10 × 10–3 x 21 watt = 0.21 watt. Ans. Problem. The network is a part of a complete circuit. What is the potential difference VB – VA, when the current I is 5A and is decreasing at a rate of 103 (A/s)? Solution: For more Study Material and Question Bank visit www.crackiitjee.in

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In accordance with law of potential distribution, for the given network,

VA  IR  E  L

dI  VB dt

 dI  is negative. dt 

as as here I is decreasing  

So VB – VA = – 5 × 1 + 15 – 5 × 10–3 (–103) i.e., VB – VA = – 5 + 15 + 5

ee .in

= 15 V Ans. Problem.

The equivalent inductance of two inductors is 2.4 H when connected in

itj

parallel and 10 H when connected in series. What is the value of

cr ac ki

inductances of the individual inductors? Solution:

As inductances obey laws similar to 'grouping of resistances', L1 + L2 = 10 H and

L1L 2  2.4 H L1  L2 

Substituting the value of (L1 + L2) from first expression into second, L1L2 = (2.4)(L1 + L2) = 2.4 × 10 = 24 so that (L1 — L2)2 = (L1 + L2 )2 – 4L1 L2 i.e., L1 – L2 = [(10)2 – 4 × 24]1/2 For more Study Material and Question Bank visit www.crackiitjee.in

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= 2H and as L1 + L2 = 10 H, L1 = 6 H and L2 = 4 H Ans. Problem. The current in a coil of self-inductance 2.0 henry is increasing according to I = 2 sin t2 A. Find the amount of energy spent during the period

.in

when the current changes from 0 to 2 A.

ee

Solution:

When the current in the coil is changing, work done in time dt

ac ki

itj

dW = P dt = eI dt

dI  I dt dt

cr

=L

dI   as e  L dt   



So W = L I dI 2

=2

 I dI 0

2

2 i.e., W = I   4J Ans. 0

Problem. For more Study Material and Question Bank visit www.crackiitjee.in

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Two different coils have self-inductances L1 = 8 mH and L2 = 2 mH. At a certain instant the current in the two calls is increasing at the same constant rate and the power supplied to the two coils is the same. Find the ratio of (a) induced voltage (b) current and (c) energy stored in the two coils at that instant. Solution: As,

eL

=

e1 L1  e2 L 2

ee .in

So,

dI dt

8 mH 4 2 mH

itj

(a)

(b)

As, P = eI

cr ac ki

dI   as dt  constt. Ans.  

= constt. [given] So, (c)

I1 e2 1   Ans. I2 e1 4

As energy stored in a coil,

 1

2

U =   LI  2

U L I  So, 1  1  1  U2 L 2  I2 

2

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2

8  1 1 =    Ans. 2  2 4 Problem. A coil of inductance L = 50 × 10–6 henry and resistance = 0.5  is connected to a battery of emf = 5.0 V. A resistance of 10  is connected parallel to the coil. Now at some instant the connection of the battery is switched off. Find the amount of heat generated in the coil after switching off the battery.

ee .in

Solution:

2

cr ac ki

1  V EL = L   2 r

1 2 Li 2 0

itj

Total energy stored in the inductor =

 Fraction of energy lost across inductor = EL .

r R  r 

LV2 = 2r R  r 

50  106  52 = 2  0.5 10  0.5 = 1.19 × 10–4 J Ans. Example.

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Let us consider a closed conducting loop is moving in a uniform magnetic field, which is perpendicular to plane of the loop. Discuss about induced emf in the loop. Solution: Method I: (a)

The magnetic flux through the loop is B = BA. Till the loop remain entirely inside the field, the flux will not change and so the net induced emf in the loop will be zero.

ee .in

Method II: We can split the loop into its four sides. The induced emf across side PQ and SR is zero, while induced emf across PS and QR is e = Bv. For the

Now let us consider closed conducting loop coming out from the

cr ac ki

(b)

itj

closed loop PQRS, the net emf becomes, e – e = 0.

magnetic field. When one of its sides QR comes out of the field, the emf across PQ, QR and RS are zero. But there is induced emf across PS, which is e = Bv. So the net induced emf across the loop is also e. The direction of induced current in the loop will be clockwise. (c)

Consider the closed conducting loop coming out of the uniform magnetic field. But velocity vector is perpendicular to one of its diagonals. The induced emf across the loop e = Bvy. As the loop is moving to the right, y increases and therefore induced emf increases and becomes maximum emax = Bv × (PR). There after induced emf starts decreasing and becomes zero when entire loop comes out of the field.

(d)

In the case, the induced emf across the ends of the conductor e = Bv



1

sin 1 

2

sin 2 

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ROTATING CONDUCTOR 1.

Rod rotating in uniform magnetic field Consider a rod of length . is rotating about an axis passing through one

B. Induced emf across the element

= B(x)dx Emf across the entire rod



itj

e = B x dx

ee .in

(de) = Bvx (dx)

cr ac ki

0

e = VP – VQ

B = 2 2.

2

Cycle wheel

Flux cutting by each metal spoke is same. Each spoke becomes cell of

B emf e = 2

2

. All such cells are in parallel fashion, therefore enet = e.

Each point on the periphery of wheel has same potential. 3.

Faraday disc dynamo: A metal disc can be assumed to be made up of number of radial

BR 2 . All conductors. The emf induced across each conductor is e = 2 such conductors behave like a number of cells in parallel. Therefore For more Study Material and Question Bank visit www.crackiitjee.in

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enet = e =

BR 2 . 2

NOTE: The induced emf in a rotating conductor does not depend on the shape of the conductor. Example.

ee .in

A conducting rod PQ is rotated in a magnetic field about an axis passing through O. The one end of the rod is at a distance a and other end is at a distance b from O. Find induced emf across the ends of the rod.

itj

Solution:

cr ac ki

The induced emf across the element of length dx is de = Bvx dx

= B(x) dx.

The emf across the whole rod b



e = B x dx a

=



B b2  a2 2

 Ans.

INDUCED ELECTRIC FIELD Consider a metal ring is placed in a uniform external magnetic field. The field extends upto cylindrical region of radius R. If we increase the For more Study Material and Question Bank visit www.crackiitjee.in

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intensity of the field, the magnetic flux through the ring will then change and by Faraday's law an induced emf and thus induced current will flow in the ring. If there is a current in the metal ring, an electric field must be present at various points within the ring, and it must have been produced by the changing magnetic flux. This induced electric field E n is just as real as an electric field by static charges. Thus we can say that changing magnetic field produces an electric field. If we replace the metal ring by a hypothetical circular path of radius r

dB , the electric dt

.in

and magnetic field is increasing at the constant rate

field induces at various points around the circular path. Due to symmetry

ac ki

Another form of Faraday's law

itj

ee

it will be tangent to each point of the path.

If En is the induced electric field, then work done by the force of this

cr

field in moving a test charge q0 around the path is (q0 E) (2r). Suppose e is the induced emf, then work done on the test charge in one revolution is eq0. Thus we can write eq0 = (q0 E n) (2r) or e = E n (2r) More generally En (2r) can be written as e=

 E.d

According to the Faraday's law e= 

dB , dt For more Study Material and Question Bank visit www.crackiitjee.in

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 (a)

E

n

.d  

dB dt

If the magnetic field increases at a steady rate, a constant induced current appears in the metal ring.

(b)

Induced electric fields appear at, various points even the ring is removed.

Magnitude of En We have;

For r < R:

or En =  (2)



dt



cr ac ki

En × 2r = 

d B  r2

itj

(1)

dB dt

ee .in

 En .d  

r dB 2 dt

For r > R :

En  2r  



d B  R 2



dt

R 2  dB  or En =  2r  dt 

Difference between electric potential and induced emf Electric field can be produced by static charge or by changing magnetic field. And electric field produced by any means exert forces on charged For more Study Material and Question Bank visit www.crackiitjee.in

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particles. The electric field produced by static charges never form closed loop, but induced electric fields form closed loop. The field produced by a static charge is of conservative nature while induced electric field is of non-conservative nature, and so its line integral over a closed path is nor zero. Thus we can say that electric potential has meaning only for fields produced by static charges; it has no meaning for induced electric field.

NOTE: Lenz's law is consistent with the principle of conservation of energy.

2.

The induced emf in Faraday's law does not have a chemical or

e. in

1.

electrostatic origin.

je

The work done in carrying a charge around the closed loop is not zero.

 0.

cr ac

 E.d

ki it

If En is the induced electric field associated with an induced emf, then

Thus here induced electric field is not a conservative field.

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