Emi

Vidyamandir Classes

Electromagnetic Induction

Electromagnetic Induction MAGNETIC FLUX

Section - 5

The magnetic flux through a circuit is a measure of the number of lines of force passing through the circuit when it is placed in an external magnetic field. Mathematically, we define flux as follows :

  B .d A where B : magnetic field vector

A : area vector of loop

If the external field is uniform, then :

B. A 

 = B A cos 

where  : angle betwen B and A .

Units : The unit of magnetic flux is Weber (Wb). 1 Wb = 1 T m2.

Faraday’s Laws of Electromagnetic Induction : According to Faraday’s Law, whenever the magnetic flux through a circuit changes, an emf is induced in the circuit. The magnitude of the induced emf is equal to the rate at which flux changes with time. Magnitude of the induced emf =

d dt

If  represents the flux through a single turn, and the loop has N turns, then Induced emf = N

d dt

This induced emf creates an induced current in the circuit whose magnitude is given as

i

induced emf 1 d  net resistance of circuit R dt

The direction of induced current (and hence that of induced emf) is obtained by Lenz’s Law.

Lenz’s Law : According to Lenz’s Law, the induced current always opposes the flux change that creates it. In other words, if the flux decreases, then the induced current tries to support the existing magnetic field and if the flux increases, the induced emf tries to balance the existing magnetic field. Self Study Course for IITJEE with Online Support

Section 5

21

Vidyamandir Classes


Induced Charge Flow : When a current is induced in the circuit due to the flux change, charge flows through the circuit and the net amount of charge which flows along the circuit is given as : q   i dt 



q

1 d 1 dt  dt R

R

|  | R

and

qN

 d |  | for N turns. R

Heat dissipation : The heat dissipated due to induced current H   Vi dt   i

d dt   i|d | dt

If the induced current is kept constant, then heat dissipated  i  |d  |  i|  |

Illustration - 19 A conducting rod AB makes contact with the metal rails AD and BC which are 0.5 m apart. If uniform magnetic field of induction 0.5 Wb/m2 acts perpendicular to the plane of the figure., (a) Calculate the magnitude and direction of the induced emf when the rod is moving to the right with speed 4 m/s. (b) If the resistance of the circuit ABCD is 0.2 ohm, calculate the force required to maintain the motion assuming smooth contacts. (c) Compare the rate at which mechanical work is done by the force with the rate of heat dissipated in the circuit. SOLUTION : (a) Let AB = l and AD = BC = x at some instant. 

= 0.5 × 5 × 0.5 = 1.25 N towards left.

 =Blx

d

 Bl

dx

 Bl v



induced emf =



induced emf = 05  0.5  4 = 1 volt.

dt

dt

(b) Induced current = i = induced emf/R = 1/0.2 = 5 A 22

Section 5

magnetic force acting on AB = B i l The applied force to maintain constant velocity must be of same magnitude but opposite direction i.e., towards right. (c) Rate of work done = Fv = 1.25 × 4 = 5 Watts. Rate of heat dissipation in resistance i2 R Self Study Course for IITJEE with Online Support

Vidyamandir Classes


= 52 × 0.2 = 5 Watts. Hence the work done by force applied to the rod is completely converted to heat in the resistance.

Illustration - 20 Space is divided by line AD into two regions. region - I is field free and region - II has a uniform magnetic field B directed into the plane of the paper. ACD is a semicirular conducting loop of radius r with centre at O, the plane of loop being in the plane of the paper. The loop is now made to rotate with a constant angular velocity  about an axis passing through O and perpendicular to the plane of the paper. The effective resistance of the loop is R. (i) Obtain an expression for the magnitude of the induced current in the loop. (ii) Show that the direction of current when the loop is entering into region II. (iii) Draw a graph between the induced current in the loop and time t. SOLUTION : (i) Flux through the loop is  = B (1/2 r2  ) at the instant when AD has rotated through angle  . Induced current I=



1

d

R

dt



I

B r 2 d

(iii) After half cycle rotation, the flux starts decreasing and hence current takes the clockwise direction. The direction of the induced current reverses after every half cycle i.e. after every  / seconds.

2 R dt Br 2  2R

(ii) As the flux is increasing when A is entering into region - II, the induced field must oppose the existing field i.e., it should be outward. Hence the induced current must be anti-clockwise.

Illustration - 21 A conducting rod AB of length l slides smoothly on two metal rails placed parallel to a long wire carrying current I as shown. A resistance R is connected between two ends of the rails as shown. The velocity of the rod is kept constant at v by applying a force. (a)

Find the induced current in the resistance.

(b)

What force has to be applied on the rod to keep the speed constant ?

Self Study Course for IITJEE with Online Support

Section 5

23

Vidyamandir Classes

Electromagnetic Induction SOLUTION : (a) Let us first calculate the flux through the circuit. Let d = flux through an infinitesimal area element of width dr at a distance r from wire. d =

0 I



xdr

2 r

where x = instantaneous position of rod al



i

   d 



0 I  dr 2 r

a

1 0 I a  l  dx  log   R 2 a  dt 

0 I v

a  l log   2R  a 

(b) Let dF = force on an infinitesimal element of width dr at a distance r from wire. 0 I dF  i dr 2r al

 Fapplied   dF 



induced current =

1

d

R

dt

i dr

2 r

a



0 I

v  0 I  a  l  Fapplied   log   R  2  a 

2

[Substitute i from part (a)]

Illustration - 22 A rectangular loop of N turns of area A and resistance R rotates at a uniform angular velocity  about Y-axis. The loop lies in a uniform magnetic field B in the direction of X-axis. Assuming that at t = 0, the plane of the loop is normal to the lines of force, find an expression for the peak value of the emf and current induced in the loop. What is the magnitude of torque required on the loop to keep it moving with constant  ? SOLUTION : As  is maximum at t = 0, after time t :

 (t) = NBA cos t magnitude of induced emf  d / dt = BAN | sin t | magnitude of induced current

BA  N =

24

(a) Peak value of emf = BA N peak value of induced current = BA N/R (b) Power input = heat dissipation per sec    = I 2 R 

 =

B 2 A2  N 2

sin  t

R

sin  t

R

Section 5


Vidyamandir Classes


Illustration - 23 A flexible circular loop 10 cm in diameter lies in a magnetic field of 1.2 T, directed into the plane of the diagram. The loop is pulled at the points indicated by the arrows, forming a loop of zero area in 0.2 sec. (a) Find the average induced emf in the circuit. (b) What is the direction of current in R. (c) What amount of charge will flow through R during this time if R = 10 ohms (neglect the resistance of the loop). SOLUTION : i = BA cos0 = BA f = 0

(b) As the flux decreases, induced field must support the existing field i.e., inwards. Hence current must be clockwise.

| f – i | = 0  BA = BA (a) Average emf induced

(c) Charge = 

|  |



time

BA t



1.2    0.05 

|  | R

2

 0.047 Volts



BA



1.2    0.05 

R

2

10

= 9.42 10–4 C

0.2

Illustration - 24 A magnetic flux through a stationary loop with a resistance R varies during the time interval  at = ( – t). Find the amount of heat generated in the loop during that time. The inductance of the loop is to be neglected. SOLUTION : 2

 2 a   2t  a   2t  4a 2  3 |d  / dt| 2  i  . Heat   i R dt   R dt R 3 R R R2 0


Section 5

25

Vidyamandir Classes


MOTIONAL EMF

Section - 6

When a conductor moves in a magnetic field, it cuts the magnetic lines of force and an emf is hence induced across its ends which is known as motional emf. If a conductor cuts the lines of force in uniform magnetic field, charges in the conductor experience a force q (v  B ) and opposite polarities are formed by accumulation of negative charges at one end. In steady state, qE  q v  B  0 where E = induced electric field 

E  v  B Q

VQ – VP = motional emf =   E . dl   (v  B ) . dl P

(i) As a particular example, let us consider a straight conducting rod of length l moving with a speed v in uniform magnetic field B and cutting the lines normally. 

induced emf =

 (v  B ) . dl

  B v dl  Bvl

(ii) Consider a rod of length rotating l with a constant angular velocity  about one of its ends. A uniform magnetic field B normal to the plane of rotation of rod exists. Emf induced across the ends of an element of dx at a distance x from the centre of rotation = B (dx) x  (where x is the speed of that element) l

Net emf across ends of rod =  B  dx  x   0

B  2 2

Illustration - 25 In the figure shown, a copper rod moving with a velocity v parallel to a long straight wire carrying a current I is shown. Calculate the induced emf in the rod, assuming that v = 5.0 m/s, I = 100 amp, a = 1.0 cm and b = 20 cm. SOLUTION :

= B (x) v dx

Motional emf induced across an infinitesimal element of length dx located at a distance x from the wire 26

Section 6


Vidyamandir Classes


Net emf across the rod =  B  x  v dx b

=

0 I

 2 x vdx a

=

0 Iv b log 2 a

IN-CHAPTER EXERCISE - D 1.

2.

3.

A search coil used to measure the magnetic fields in to be made with a radius of 2 cm. It is to be designed so that flipping it 180 in a field of 0.1T causes a total charge of 1  10–4 C to flow in a charge measuring instrument when the total circuit resistance is 50 ohm. How many turns should the coil have ? The current in the wire AB of infinite length in given figure is upward and is increasing steadily at a rate dI/dt. (a) At an instant when the current is I. What are the magnitude and direction of the magnetic field B at a distance r from the wire ? (b) What is the flux d through the narrow shaded strip ? (c) What is the total flux through the loop ? (d) what is the induced emf in the loop ? (e) Evaluate the numerical value of the induced emf if a = 10 cm, b = 30 cm, l = 20 cm, dI/dt = 2A /s. A uniform magnetic field of induction B is changing in magnitude at a constant rate dB/dt. You are given a mass m of copper which is to be drawn into a wire of radius r and formed into a circular loop of radius R. If  is the specific resistance and  is the density of the wire, show that the induced current in the loop does not depend on the size of the wire or of the loop and , assuming B perpendicular to the loop, is given by i 

m dB 4 dt

4.

A rod of length d and a rectangular loop of width d are released together and fall in a uniform field, as shown in figure. Is there any difference in their motion ? (Assume the loop is never totally within the field.)

NOW ATTEMPT IN-CHAPTER EXERCISE-D FOR REMAINING QUESTIONS IN THIS EBOOK Self Study Course for IITJEE with Online Support

Section 6

27

Vidyamandir Classes


INDUCTANCE

Section - 7

Whenever the current through a circuit changes, flux due to that current also changes. An emf is hence induced and is known as self inducted emf because there is no external field but only the field due to the circuit’s own current that is responsible for the flux change. This process is known as self induction. If i is the current in the circuit, then the flux due to that current is  i =Li where L is known as the self-induction of the circuit and its unit is Henry (H).

d Self induced emf =

L

dt

di dt

Energy stored in a circuit if current changes from 0 to i =  E i dt   L i

=  L i di  0



U

1

1

di

i dt

dt

Li 2

2

L i2

2

Self Induction of a Coil : Consider a coil of N turns and area of crossection A carrying a current i. The length of the coil is  ( A). 



 0 i N   NBA N   A  l 



0 N 2 A

i

l Comparing with  = L i, we get : L

0 N 2 A l

28

Section 7


Vidyamandir Classes


Combination of Inductors (i)

Series combination :

V1  L1

dI dt

V2  L2 V L

dI dt

V3  L3

dI dt

dI dt

V XY  V1  V2  V3 L

dI dI dI dI  L1  L2  L3 dt dt dt dt

L  L1  L2  L3 In general Leff   Li in series (ii)

Parallel Combination :

dI1 dI dI  L2 2  L3 3 dt dt dt I  I1  I 2  I3

V XY  L1

and V XY  L

dI dt

dI dI1 dI 2 dI3    dt dt dt dt V V V V    L L1 L2 L3

(V  V XY )


Section 7

29

Vidyamandir Classes


1 1 1 1    L L1 L2 L3 In general

1 1  in parallel. L Li

RL CIRCUIT : (a)

Growth of Current :

Let us connect a coil of self-induction L with a resistance R across a cell of emf E as shown. If the switch S is thrown in contact with point 1 at t = 0, current i in the circuit tends to grow. Hence an emf is induced across the coil in such a direction as to oppose this current as shown. By Kirchoff’s voltage law, we have : di E–iR – L =0 dt di dt   E  iR L i



 R di

t



 E  iR  0



 R dt

0

log  E  iR  

L

Rt i  0 L



i t  

E

1  e Rt / L   R

The current grows exponentially with time and tends to attain the steady state value E/R. The time after which current has grown upto almost 63 % of steady state value is known as time constant  . Here  

L

and

R

(b)

i   

E  1 E 1    0.63 R e R

Decay of Current : Let i0 be the steady state current in the coil and the switch at that instant is thrown in the contact with point 2 so that there is no cell in the circuit. The current, hence tends to decrease and an emf is induced across L to support this current.

di     L   iR  0 dt   30

Section 7


Vidyamandir Classes


Note that di/dt is negative quantity i



 0

t

di



i

 0

 R dt L

i = i0 e–Rt/L



The current decays exponentially with time.

LC Circuit : An ideal capacitor fully charged is connected to an ideal inductor and switch is closed at t = 0.

From Kirchoffs low L

dI q  0 dt C

d 2q dt 2 d 2q 2



1 q0 LC

2 

1 LC

  2q  0

dt Solution of equation (similar to SHM) is

q  q0 cos (t   ) i

dq   q0 sin(t   ) dt

t 0 

I 0



 0

q  q0 cos t i  i0  sin t

At any time t VC 

q q0  cos t C C

VL  L

dI   2 Lq0 cos t dt


Section 7

31


UB 

1 2 1 LI  L 2 q 2 sin 2  t 2 2

UC 

q 2 q02 cos 2 t  2C 2C

Vidyamandir Classes

q02 Total energy  U B  U C  which is constant 2C

This system can be compared with a spring mass system is SHM. x  x0 cos( t   ) and can be analyzed accordingly.

Illustration - 26 A series circuit having a self-induction 3 H, a pure resistor 10 ohm and a source of constant voltage 3 V is closed at time t = 0. Find : (a) time constant of the circuit. At the end of 1 time-constant, find : (b) the current in the circuit, (c) the rate at which the current in the circuit is increasing, (d) the rate at which joule’s heating is produced, (e) the rate at which energy is stored in the magnetic field, (f) the rate at which energy is delivered by the source. Is the total energy conserved ?

32

Section 7


Vidyamandir Classes


SOLUTION : (f) Rate at which cell delivers energy = E i = 3 (0.19) = 0.57 Watts.

i = 3/10 (1 – e–10 t/3) (a)

 = L/R = 0.3 sec

(b)

i (0.3) = 0.3 (1 – 1/e) = 0.19 A

(c)

di/dt = e–10t/3 = 1/e = 0.37 A/s

(d)

P = i2R = (0.19)2 10 = 0.361 Watts

dU (e)

It is earily seen that E i = i2 R + L i

di dt

d 1 2 di  Li   Li dt dt  2 dt  = 3  0.19  0.37 = 0.21 W 

MUTUAL INDUCTION : Consider two coils C1 and C2 placed as shown. By varying current i1 in coil C1, we change the flux not only through C1 but also through coil C2. The change in flux 2 through C2 (due to change in current i1) induces an emf in the coil C2. This emf is known as mutually induced emf and the process is known as mutual induction.

2  i1 

 2 = M i1

where M is called as mutual inductance of the pair of coils. The coil C1 in which i varies is often called primary coil and the coil C2 in which the emf is induced is called secondary coil. induced emf in coil C2 = E2

E2 

d2 dt

M

di1 dt

The mutual inductance is maximum when the coils are wound up on the same axis. It is minimum when the axes of coils are normal to each other.


Section 7

33

Vidyamandir Classes


Illustration - 27 A long solenoid of length 1 m, cross sectional area 10 cm2, having 1000 turns has wound about its centre a small coil of 20 turns. Compute the mutual inductance of the two circuits. What is the emf in the coil when the current in the solenoid changes at the rate of 10 Amp/s ? SOLUTION : Let

N1 = number of turns in solenoid ; N2 = number of turns in coil

A1 and A2 be their respective areas of crossection. (A1 = A2 is this problem)

Comparing with 2 = M i1, we get : Mutual inductance = M 

Flux 2 through coil created by current i1 in solenoid is 2 = N2 (B1 A2)

 0 i1 N1  2  N2   A2 l  

M

0 N1 N 2 A2 l

4   107  1000  20  10  104 1

= 2.51 ×

10–5

H

Magnitude of induced emf = E2  M

di1 dt

E2 = 2.51  10–5  10 = 2.51  10–4 V.

 0 N1 N 2 A2  2    i1 l  

Illustration - 28 (a) Calculate the mutual inductance between two coils when a current of 4 amp. changes to 12 amp. in 0.5 sec. and induces an e.m.f. of 50 milli-volts in the secondary. (b) Also calculate the induced e.m.f. in the secondary if current in the primary changes from 3 amp. to 9 amp. in 0.02 secs. SOLUTION :

di (a)

dt



12  4

 16 amp / sec

(b)

0.5

di



dt

 di  Induced emf = M    dt  50  103    M  = 3.125 × 10

Now

93

 300 amp / sec

0.02 emf = (3.125 × 10–3) (300) = 0.9375 volt

–3

Henry..

16

34

Section 7


Vidyamandir Classes


Illustration - 29 Calculate the length of a thin wire of uniform cross section to manufacture a solenoid of length l0 = 100 and the self-induction 1 mH. (Assume the cross section diameter of solenoid to be negligible with respect to its length). SOLUTION : 2

L

   rN  0 N 2 A  N 2 r 2 . Calculate rN taking l = 1m and L = 10–3 H..  0  0 l l l

Length of wire required = 2rN.

Illustration - 30 A coil of 100 turns and 1 cm radius is kept coaxially within a long solenoid of 8 turns per cm and 5 cm radius. Find the mutual inductance. SOLUTION : The magnetic field B in the primary of solenoid The mutual inductance is therefore, given by N s M  0 N p N s As B = m0 Np i i The magnetic flux linked with the secondary coil Substituting the given values, we get : is given by Ns  = Ns B As = Ns (m0 Np i) As where As is the cross sectional area of secondary of solenoid.

M = (4× 10–7) weber/amp–m) (8000/m) × 100 × (× 10–7m2) = 3.15× 10–5 henry.

Illustration - 31 A long solenoid having 1000 turns per cm carries an alternating current of peak value 1 ampere. A search coil having a cross-sectional area of 1× 10–4 m2 and 20 turns is kept in the solenoid so that its plane is perpendicular to the axis of the solenoid. The search coil registers a peak voltage of 2.5× 10–2 volt. Find the frequency of the current in the solenoid. SOLUTION : We know that the magnetic field of the solenoid is given by B =  i N,

. . . . (i)

where i is the current and N is the number of turns per metre. The magnetic flux-linked with the search coil is given by  = B A Ns . . . . (ii) where A = area of search coil and Ns is the total number of turns in the search coil. Self Study Course for IITJEE with Online Support

Substituting the value of B from eq. (i) eq. (ii), we get :  = (0 i N) A Ns

d Now

dt

 0 A N N s

di dt

where i (t )  i0 sin t d = 0 A N Ns (i0  cos  t) dt

. . . . (iii)

Section 7

35

Vidyamandir Classes

Electromagnetic Induction The peak value is given by

Solving for , we get : = 99.47 per sec.

 d   0 A N N s i0     dt  peak

 Substituting the given values, we have

f 

 2



99.47

 15.8 / sec

2  3.14

2.5× 10–2 = (4× 10–7) × (1× 10–4) × (105) × (20) × (20) × 1 × 

IN-CHAPTER EXERCISE - E 1.

Two coils are wrapped around cylindrical forms, as shown in figure. One coil is in series with a battery, a switch, and a variable resistance. The other is connected to an ammeter. State the direction of the induced current measured by the ammeter (x to y, or y to x) in the following circumstances : (a) when the switch is first closed ; (b) with the switch closed, the resistance is decreased (c) with the switch closed, the coils are moved apart.

2.

Two loops lie side by the side on a table. If a clockwise current suddenly starts to flow in one, what is the sense of the induced current in the second ?

3.

Can there be an induced emf in an inductor even if the current through it is zero ?

4.

A real inductor has some resistance. Can be potential difference across the terminals of the inductor be (a) greater than the induced emf ; (b) less thant the induced emf ?

5.

The self-induced emf in a solenoid of length 25 cm and radius 1.5 cm is 1.6 mV. When the current is 3A and increasing at the rate 200 A/s. (a) What is the number of turns ? (b) What is the magnetic field within the solenoid at the given instant ? Ignore other effects.

NOW ATTEMPT IN-CHAPTER EXERCISE-E FOR REMAINING QUESTIONS IN THIS EBOOK

36

Section 7


Vidyamandir Classes


INDUCED EMF AND ELECTRIC FIELDS

Section - 8

A changing magnetic field gives rise to an induced electric field. This electric field created in the conductor as a result of changing flux is non conservative, time varying and is mutually perpendicular to the magnetic fields. Hence the electric field lines representing the induced electric field form closed curves. A conducting loop of radius r is kept in a time varying magnetic field.

 

d B dt

The induction of current in the loop implies presence of induced electric field. The work done in moving the   test charge along the loop is qE (2 r ) or E  2 r Hence we can write   d  E  ds   dtB

Illustration - 32 A uniform magnetic field of induction B fills a cylindrical volume of radius R. A rod PQ of dB . Find the emf produced by the changing magnetic length 2 is placed as shown. If B is changing at a rate dt field that acts between the end of the rod.

SOLUTION : Consider a circle of radius r ( R)


Section 8

37

Vidyamandir Classes

Electromagnetic Induction d B  d   r2B dt dt dB    r 2 dt



 

E 2 r   r 2



dB dt

1 dB E   r2 2 dt AB  d  dW  ( E cos  )d  AN  rd  d  cos 



 1 dB  dW   r  rd  2 dt  1  dB dW   r 2 d  2  dt

Area of OAB 

1 2 r d 2

Total work done from P to Q = Area OPQ 

38

 PQ 

dB 1   2 R 2   2 dt 2

 PQ 

dB   R2  2 dt

Section 8

dB dt


Vidyamandir Classes


SUBJECTIVE SOLVED EXAMPLES Example - 1 Two long parallel horizontal rails, a distance d apart and each having a resistance  per unit length, are joined at one end by a resistance R. A perfectly conducting rod MN of mass m is free to slide along the rails without friction. There is a uniform magnetic field of induction B, normal to the plane of paper and directed into the paper. A variable force F is applied to rod MN, such that as the rod moves, a constant current I flows though R. (i) Find the velocity of the rod and the applied force F as a function of distance x of the rod from R. (ii) What fraction of work done per second by F is converted into heat ? SOLUTION : (i) Instantaneous flux =  = B x d 

1 d induced current  R  2  dt

F (x) = B I d +

2 mI 2  B2d 2

 R  2x  

using (i) (ii) (Work done per sec by F) = F v Heat dissipated per sec = I2 (R + 2 x  ) 2 Heat dissipated per sec I  R  2 x   work done per sec Fv

Bvd I R  2 x



I v(x) = (R + 2 x ) Bd

=

I 2  R  2x   I 2 R  2 x 

. . . .(i)

2m  I 3 3

B d

3

 R  2 x  2

B3d 3

Magnetic force acting on the rod = B I d dv  F – B I d = mv dx

= 3 3 B d  2m  I  R  2 x  


Subjective Solved Examples

39

Vidyamandir Classes


Example - 2 A horizontal copper conductor of mass m slides down two smooth parallel vertical copper bars, due to gravity. At the top, the bars are connected through a resistor of resistance R. The distance between two bars is l and the system is placed in a vertical uniform horizontal magnetic field of induction B directed normal to the plane of the loop so formed. Find the maximum velocity attained by the copper connector, assuming bar to be smooth and of negligible resistance and neglecting self-induction of the loops. SOLUTION : Let x be the instantaneous position of the connector PQ from the top. The flux through the circuit is  = B l x  induced current 1 =I= R

d dt

Blv  R

From the force diagram : mg – Bil = m dv/dt

B2 l 2 v dv mg  m . . . .(i) R dt dv/dt decreases and v increases as the connector PQ comes down.

  

By taking t  , dv/dt  0, we get the terminal velocity.

Mg 

B2 l 2 v 0 R

 terminal or maximum velocity = v 

B2 l 2 v  magnetic force on PQ = B i l = R By Lenz’s law, the induced current is clockwise and magnetic force tends to retard the connector. By using F  i (l  B), we get the direction of force B i l.

mg R B2 l 2

Note that we can get the expression of velocity v as a function of time by solving the differential equation (i).  B2l 2  t mg R  V t   1  e mR   B2 l 2    

Example - 3 A wire bent as a parabola y = ax2 is located in a uniform magnetic field of induction B, the vector B being perpendicular to the plane X-Y. At the moment t = 0 a connector starts sliding translation wise from the parabola apex with a constant acceleration  . Find the emf of electromagnetic induction in the loop thus formed as a function of y. SOLUTION : Let P : (x, y) induced emf = Bvl  B 2wy  2 x   B 2 wy 2

y a

[Note that for constant acc., v2 = 02 + 2wy]

40



Vidyamandir Classes


Example - 4 A square frame with side a and a long straight wire carrying a current I are located in the same plane as shown in figure. The frame translates to the right with a constant velocity v. Find the emf induced in the frame as a function of distance x. SOLUTION : xa

   d 

 x

emf 



0 I  Ia xa . adr  0 log 2 r 2 x

 Ia d x   a  dx  0 . . 2  dt 2 x  a  x  dt  0 Ia 2 v 2 xx  a

(Negative sign indicates that  is decreasing) Alternatively, emf induced = B1va  B2va

0 I  1 1  = (B1  B2) va = 2  x  x  a  va  

Example - 5 A conducting rod AB of mass m slides without friction over two long conducting rails separated by a distance l. At the left end the rails are interconnected by a resistance R. The system is located in a uniform magnetic field perpendicular to the plane of the loop. At the moment t = 0 the rod AB starts moving to the right with an initial velocity v0. Neglecting the resistances of the rails and the rod AB, as well as the self - inductance, find : (a) the distance covered by the rod until it comes to a standstill ; (b) the amount of heat generated in the resistance R during this process. SOLUTION :

Let

dx  v0e kt dt i

B 2l 2v mdv  R dt

Bv  mdv and bil  R dt 2 2  v  V0e  B l / mR

B 22 k mR 

x



0

v mR v0 v0e kt  0  k B 2l 2

1 mv02 2 Also try finding out heat dissipation by

Heat dissipation = loss in KE 

 2 i R dt 0



and check if answer comes out to be same.



NOW ATTEMPT OBJECTIVE WORKSHEET TO COMPLETE THIS EBOOK

41

Vidyamandir Classes


THINGS TO REMEMBER [B] Electromagnetic Induction & AC Circuit 1.

Magnetic Flux The magnetic flux through a circuit is a measure of the number of lines of force passing through the circuit when it is placed in an external magnetic field. Mathematically, we define flux as follows :

   B d A where B : magnetic field vector If the external field is uniform, then :

A : area vector of loop

  BA  2.

where  : angle between

  BA cos 

Faraday’s Laws of Electromagnetic Induction : According to Faraday’s Law, whenever the magnetic flux through a circuit change, an emf is induced in the circuit. The magnitude of the induced emf is equal to the rate at which flux changes with time. Magnitude of the induced emf 

d dt

This induced emf creates an induced current in the circuit whose magnitude is given as

i

induced emf 1 d  net resistance of circuit R dt

3.

Len’s Law : According to Len’z Law, the induced current always opposes the flux change that creates it. In other words, if the flux decreases, then the induced current tries to support the existing magnetic field and if the flux increases, the induced emf tries to balance the existing magnetic field.

4.

Induced Charge Flow : When a current is induced in the circuit due to the flux change, charge flows through the circuit and the net amount of charge which flows along the circuit is given as :

q   idt  



42

q

1 d 1 dt   d R dt R

  and q  N R R

Things to Remember

for N turns.


Vidyamandir Classes

5.


Heat dissipation : The heat dissipated due to induced current H   vi dt   i

d dt   i d dt

If the induced current is kept constant, then heat dissipated  i  d  i  6.

Motional EMF When a conductor moves in a magnetic field, it cuts the magnetic lines of force and an emf and an emf is hence induced across its ends which is known as motional emf.

vQ  vP  motional emf   v  B  dl (a)

(b)

As a particular example, let us consider a straight conducting rod of length l moving with a speed v in uniform magnetic field B and cutting the lines normally. Consider a rod of length rotating with a constant angular velocity  about one of its ends. A uniform magnetic field B normal to the plane of rotation of rod exists. Net emf across ends of rod 

7.

B  2 2

Inductance Whenever the current through a circuit changes, flux due to that current also change. An emf is hence induced and is known as self inducted emf because there is no external field but only the field due to the circuit’s own current that is responsible for the flux change. This process is known as self induction. If i is the current in the circuit, then the flux due to that current is  i

  L i where L is known as the self-induction of the circuit and its unit is Henry (H). Self induced emf 

8.

d di L dt dt

Energy stored in a circuit if current changes from 0 to i U

1 2 Li 2


Things to Remember

43

Vidyamandir Classes


9.

Self induction of a Coil : Consider a coil of N turn and area of crossection A carrying a current i. The length of the coil is l (l  A ).

10. RL CIRCUIT : (a)

Let us connect a coil of self-induction L with a resistance R across a cell of emf E as shown. If the switch S is thrown in contact with point 1 at t = 0, current i in the circuit tends to grow. i (t ) 

E 1  e Rt / L R





The current grows exponentially with time and tends to attain the steady state value E/R. The time after which current has grown upto almost 63% of steady state value is known as time constant  . Here



E  1 E L and i ( )  1    0.63 R  e R R

11. Decay of Current : Let i0 be the steady state current in the coil and the switch at that instant is thrown in the contact with point w so that there is no cell in the circuit. The current, hence tends to decrease and an emf is induced across L to support this current. i  i0 e  Rt / L

The current decays exponentially with time.

44

Things to Remember


Vidyamandir Classes


12. Mutual Induction Consider two coils C1 and C2 placed as shown. By varying current i 1 in coil C 1, but also through coil

C 2 . The change in flux  2 through C 2 (due to change in current i 1 ) induces an emf in the coil C 2 . This emf is known as mutually induced emf and the process is known as mutual induction.

2  i1



  M i1

where M is called as mutual inductance of the pair of coils. The coil C 1 in which i varies is often called primary coil and the coil C 2 in which the emf is induced is called secondary coil. Induced emf in coil C 2  E 2

E2 

d 2 di M 1 dt dt

The mutual inductance is maximum when the coils are wound up on the same axis. It is minimum when the axes of coils are normal to each other.


Things to Remember

45

Vidyamandir Classes


SOLUTIONS - IN-CHAPTER EXERCISE-D

2.

N |   qR  N  20 turns [ =  –  = +(–Br2) – (Br2) = –2Br2] f i R 2 B r 2 0 I 0 I  0 I  b dr n (a) (b) (c) 2 r 2 r 2 a 0 b dI n (d) (e) 8.83 × 10–8 V 2 a dt

3.

m = [A(2R)] ;

q

1.

[A = r2 ] 2

 dB / dt   R RA dB m dB emf   .  m   2 R  2  dt 4 dt  AR   resistance 2 A Bdv   There is an induced current in the loop  i   and hence it experiences an upward magnetic force. The R   loop will not fall with acceleration = g but the rod will fall with acceleration ‘g’. Loop will fall with an acceleration less than ‘g’. i

4.

SOLUTIONS - IN-CHAPTER EXERCISE-E 1.

2.

A clockwise current produces field in the other loop going into the table. To oppose the increase in flux, a clockwise current is induced in second loop. di Yes because induced emf = L dt (a) Yes, for increasing i (b) Yes, for decreasing i.

3. 4.

 N2A 1.6  103  8 106 H ; L  0 200  0 IN Solve for N and B  ( = 25 cm ; I = 3A) 

1.6  103  L

5.



46

di dt

Solutions

 L


Vidyamandir Classes

My Chapter Notes


Vidyamandir Classes

Illustration - 1


Vidyamandir Classes


Vidyamandir Classes


Emi

Recommend Documents