06.electromagnetic Induction F

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6.ELECTROMAGNETIC
INDUCTION

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Electromagnetic Induction
Introduction
Experiments performed by scientists like Oersted, Ampere etc, during the early years of 19th century established that
magnetic fields may be produced using electric currents. It was natural to enquire whether an electric current be
produced using magnetic fields. Producing electric currents from magnetic fields was not so easy as to produce
magnetic fields from currents. However, careful experimental observations made independently by Michael Faraday
in England and Joseph Henry in USA around the year 1831, showed that under certain conditions electrical current
can be produced using magnetic fields. It was found that when the magnetic flux linked with a loop of a conductor
changes there will be an electric current in the loop. The phenomenon in which an emf is developed in a conductor
due to the change in the magnetic flux linked with it is called the electromagnetic induction. Almost all the entire
electrical energy used in the world today is produced making use of electromagnetic induction.
Magnetic flux
The number of field lines passing normally across a surface is called the flux across the
surface. The flux associated with a magnetic field is defined in a manner similar to that
used to define electric flux. Let dS be an element of area on an arbitrary shaped surface 
as shown. If the magnetic field at this element is B , the magnetic flux through the element
is,
dB = B  dS = BdScos  .
dS here is a vector that is perpendicular to the surface and has a magnitude equal to the area dS and  is the angle
between B and dS . In general dB varies from element to element. The total magnetic flux through the surface is
the sum of the contributions from the individual area elements.
B =  Bdscos  =  B  dS
(i)Magnetic flux is a scalar quantity (dot product of two vector quantities is a scalar quantity)
(ii) The SI unit of magnetic flux is tesla–meter2 (1 T-m2). This unit is called weber (1 Wb)
1 Wb = 1 T-m2 = I N-m/A
 Thus unit of magnetic field is also weber/m2 (1 Wb/m2),
or 1 T = 1 Wb/m2
In the special case where B is uniform over a plane surface with total area S and is normal to the surface then cos   1
and B = BS
Faraday’s law of electromagnetic induction

The results of systematic experimental observations made by Faraday may be summarised in the form laws called
Faradays laws of electromagnetic induction.
These laws states that
(i) whenever the flux of magnetic field through the area bounded by a closed conducting loop changes, an emf is
produced in the loop, and
(ii) the magnitude of emf induced in a coil is equal to the rate of change of magnetic flux linked with the coil.
d
The induced emf is given by E = − ...(1)
dt
where  =  B  dS is the flux of the magnetic field through that area.
The law described by equation (1) is called Faraday’s law of electromagnetic induction. The flux may be changed
in a number of ways. One can change the magnitude of the magnetic field B at the site of the loop, the area of the
loop or the angle between the area-vector dS and the magnetic field B . In any case, as long as the flux keeps
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changing, the emf is present. The emf produced so drives an electric current through the loop. If R is the resistance
of the loop, the current then is
E 1 d
i= =− ...(2)
R R dt   
a
The emf developed by a changing flux is called induced emf and the current
produced by this emf is called induced current.
Suppose the circuit (loop) consists of N loops all of same area and if B is  v 
the flux through one loop, then the total induced emf is given by
d
e = −N B b
dt   
significance of −ve sign will be explained below.
Direction of induced current

Lenz’s law
Soon after Faraday gave his law of induction, Lenz devised a rule-now known as Lenz’s law for finding the direction
of an induced current in a loop.
An induced current has a direction such that the magnetic field due to this current opposes the change in the magnetic
flux that induces the current.
Furthermore, the direction of an induced emf is same as that of the induced current.
Motional electromotive force
(emf induced in a rod moving in a magnetic field)
(emf Until now, we considered the cases where an emf is induced   
++
in a stationary circuit placed in a magnetic field when the field ++
changes with time. We now describe what is called motional emf,   
which is the emf induced in a conductor when it is moving l
through a constant magnetic field.   

−−
The straight conductor of length l shown in figure is moving −−
through a uniform magnetic field directed into the page (denoted   
by the sign ). For simplicity let us assume that the conductor is
moving in a direction perpendicular to the field with constant velocity under the influence of some external agent.
The electrons in the conductor experience a force, Fm = −e(v  B)
Under the influence of this force, the electrons move to the lower end of the conductor and accumulate there, leaving
a net positive charge at the upper end. As a result of this charge separation, an electric field is produced inside the
conductor. The charges accumulate at both ends until magnetic force evB which is along downward direction is
balanced by the upward electric force eE. At equilibrium electrons stop moving. The condition for equilibrium
requires that, eE = evB or E = vB
The electric field produced in the conductor (once the electrons stop moving and E is constant) is related to the
potential difference across the ends of the conductor by,
V = El = Blv
 V = Blv
where the upper end is at a higher electric potential than the lower end.
Thus, “a potential difference is maintained between the ends of the conductor as long as the conductor continues to
move through the uniform magnetic field.”
Now let us suppose the moving rod slides along a stationary U-shaped conductor, forming a complete circuit. No
magnetic force acts on the charges in the stationary U-shaped conductor, but there is an electric field resulting from
the charge accumulations at a and b. Under the action of this field a counter clockwise current is established around
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this complete circuit. The moving rod acts as a source of electromotive force. Within it, positive charge moves from
lower to higher potential and in the remainder of the circuit, charge moves from higher to lower potential. We call
this a motional electromagnetic force denoted by e, we can write,
e = Bvl …(1)
If R is the resistance of the circuit, then current in the circuit is,
e Bvl
i= = …(2)
R R
Fleming’s right hand rule
The direction of induced emf in a conductor in a magnetic field is given by Fleming’s right hand rule.
If the fingers of the right hand are held such that the fore finger, middle finger and the thumb are mutually
perpendicular and the fore finger shows the direction of field and the thumb shows the direction of motion of the
conductor then the middle finger shows the direction of induced current. This principle is used in ac generator.
Induced electric field

Consider a conducting loop which is located in a magnetic field B . The free electrons cannot flow in the loop until
an electric field is applied. As long as B is constant no electric field is induced. Suppose the flux of magnetic
induction through the loop starts changing say at t = 0, then an electric field E is produced. Obviously this electric
field is produced by the changing magnetic field and not by charged particles.
d d
Using Faraday’s law of induction, induced emf ‘e’ is given by e = − or,  E  dl = − .
dt dt
A conducting closed loop need not be there to have an induced electric field E . As long as B keeps changing, the
induced electric field is present. If a closed loop is there, the free electrons start drifting and consequently an induced
current results. Changing of B is not the only method of producing an induced electric field and consequently an
induced emf. There are other methods also.
Various methods of producing induced emf
The three major methods generally employed are
(i) change of B , the magnetic field acting on the object
(ii) change of area of material
(iii) change of relative orientation of surface area and the applied magnetic field
Inductors and inductance

An inductor is a coil of wire (conductor) with a number of turns as shown in the figure.
Such an inductor can be used to produce a desired magnetic field. It is essentially a short
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solenoid. Here we consider a long solenoid (specifically, a short length near the middle of a long solenoid) as our
basic type of inductor. If a current i is established in the windings of an inductor (a solenoid), the current produces a
magnetic flux B through the central region of the inductor.
B  I or B = LI
The constant of proportionality L is called the inductance of the inductor. Then,
N B
L= … (1)
i
where N is the number of turns. The windings of the inductor share the flux, and the product NB is called the
magnetic flux linkage. The inductance L is thus a measure of the flux linkage produced in the inductor per unit of
current.
As the SI unit of magnetic flux is tesla-square meter, the SI unit of inductance is the tesla-square meter per ampere
(T. m2/A). This is called henry (H), after American physicist Joseph Henry.
Thus, 1 henry = 1 H = 1 T . m2/A.
Inductance of a solenoid
Let us consider a long solenoid of cross-sectional area A. Let us find the inductance per unit length near its middle.
Consider a length l near the middle of this solenoid. The flux linkage for this section of the solenoid with total number
of turns N is
NB = (nl)(BA)
n being the number of turns per unit length of the solenoid and B being the magnitude of the magnetic field within
the solenoid.
The magnitude B as we already know, is given by B = 0 in,
so from equation
NB (nl)(BA) (nl)(0in)(A)
L= = = = 0 n2 lA
i i i
L
The inductance per unit length for a long solenoid near its centre is = 0n2A
l
Hence one can conclude that the inductance only depends on the geometry of the device.
Self induction
L
Self induction is the property of a coil by virtue of which, the coil opposes any change
in the current flowing through it by inducing an emf in itself. For this reason, self
induction is also referred to as the inertia of electricity. In the figure, current in a coil L
is changed by varying the contact position on a variable resistor, a self induced emf
appears in the coil while the current is changing. R
With current increasing, the self induced emf (e) across the coil appears in a direction which opposes the increase.
i (increasing) i (decreasing)
i e i e
With decrease of current, the self induced emf (e) appears across the coil in a direction, such that it opposes the
decrease. Therefore, it would be in the direction of i.
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Coefficient of self induction
Suppose I is the current flowing through a coil at any time and  is the amount of magnetic flux linked with the coil
at that time.
It is found that   I or  = Li,
where L is a constant of proportionality and is called coefficient of self induction or self inductance of the coil.
The emf induced in the coil is given by
d d dI
e=− = − (LI) or e = −L
dt dt dt
dI
If = 1, then e = −L  1 or L = −e
dt
Hence coefficient of self induction of a coil is numerically equal to the emf induced in the coil when rate of change
of current through the coil is unity.
The SI unit of L is henry. Self inductance of a coil is said to be one henry (H) if a current changing at the rate of 1
ampere/sec through the coil induces an emf of 1 volt in the coil.
Energy stored in an inductor
L
In the LR circuit shown the current grows in the circuit and the magnetic field
increases in the inductor. Part of the work done by the battery during the process
is stored in the inductor as magnetic field energy and the rest appears as thermal R
energy in the resistor. After sufficient time, the current and hence the magnetic S
E
field becomes constant and further work done by the battery appears completely
di
as thermal energy. If I is the current in the circuit at time t, we have E = L + iR
dt
t t i
or, Eidt = i R dt + Lidi

2
or,  E i dt =  i R dt +  Li di
2
0 0 0
t t
1
or,  E i dt =  i 2 R dt + Li 2 … (1)
0 0
2
The first term on right-hand side of equation (1) is the total thermal energy (Joule heat) developed in the resistor in
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time t. Thus Li2 is the energy stored in the inductor as the current in it increases from 0 to i. As the energy is zero
2
when the current is zero, the energy stored in an inductor, carrying the current i, is
1
U = Li 2
2
Energy density in magnetic field

Let us consider a long solenoid of radius r, length l and with n turns per unit length. If it carries a current i, the
magnetic field within it is B = 0 ni
Neglecting the end effects, the field outside is zero. The self-inductance of this solenoid is
L = 0 n2r2l
1 1
The magnetic energy is, therefore, U = Li2 = 0 n2r2li2
2 2
2
1 B
= (0 ni) 2 V = V
20 20
where V = r2l being the volume enclosed by the solenoid. As the field is uniform throughout the volume of the
solenoid and zero outside, the energy density, is
U B2
u= =
V 20
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Self inductance of a long solenoid
A long solenoid is defined as one of length very large compared to its radius of cross-section. The magnetic field B
at any point inside such a solenoid is practically constant and is given by
 NI
B= 0 … (1)
l
0 being the absolute magnetic permeability of free space/air, which forms the core of the solenoid, l being the length
of the solenoid, and N the total number of turns in the solenoid.
 Magnetic flux through each turn of the solenoid = B  area of each turn
 N 
=  0 I  A
 l 
where A is area of each turn of the solenoid.
Total magnetic flux linked with the solenoid is therefore equal to flux through each turn  total number of turns
n
i.e.,  = 0 IAN … (2)
l
d −0 N 2 A dI
Induced emf = − = … (3)
dt l dt
dI
Comparing this equation with e = −L
dt
NA 2
We have L = 0 . In this expression 0 is replaced by  = 0r in any other material.
l
Mutual induction
When the current passing through a coil or circuit changes, the magnetic IP IS
flux linked with a neighbouring coil or circuit will also change. Hence an changing
current
emf will therefore be induced in the neighbouring coil or circuit. The Load
 P S R
phenomenon in which an emf is induced in one coil due to the changing
M
current in a neighbouring coil is called as mutual induction. The coil or
circuit in which the current changes is called primary while the other in which emf is set up is called secondary.
In case for two coils situated close to each other, flux linked with the secondary due to current in the primary, i.e., s
is given by
S  IP or S = MIP … (1)
where M is a constant of proportionality called coefficient of mutual induction or simply mutual inductance of the
pair of coils. From equation (1)
M = S if IP = 1A
That is coefficient of mutual inductance of two coils or circuits is numerically equal to the flux linked with one circuit
or coil when unit current flows through the other.
d
According to Faraday’s law of electromagnetic induction e = −
dt
dI dI
so, e = −L or eS = −M P … (2)
dt dt
 i.e., coefficient of self induction is numerically equal to emf induced in a coil when the rate of change of current in it is
unity. While coefficient of mutual induction is numerically equal to emf induced in one coil when the rate of
change of current in the other coil is unity.
 dI   dI 
1. As |e| = L   or |e| = M   , the dimensions of inductance, i.e., [L] as well as [M] is
 dt   dt 
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 dI −1   Ml2 T −2 T 
[e]    =   = [ML2T−2A−2] … (3)
  dt    AT A
And so SI unit of L as well as M is
J Vs Wb Tm 2
kg m2 s−2 A−2 = = =   s = = … (4)
A2 A A A
and is called henry (H).
2. An inductance is said to be ideal with no resistance. In practice inductance always has a
resistance, i.e., one cannot have inductance without having resistance. i.e., one can have a
resistance with or without having inductance. A resistance without inductance is called non- R  0 and L  0
inductive resistance.
Relation between mutual inductance and self inductance

The mutual inductance M of two coils or circuits having self-inductance L1 and L2 is given by
M = k L1L 2 … (1)
k being a constant called coefficient of coupling. If the coils are wound over each other the coupling is said to be
‘tight’ otherwise ‘loose’. For tight coupling k = 1 and so M = L1L 2 while for loose coupling
0 < k < 1 and hence M < L1L 2 .
Furthermore, from expression (1) it is also clear that if L = 0, M will be zero, i.e., a system cannot have mutual
inductance without having self-inductances. However, converse may or may not be true, i.e., if mutual inductance of
a system is zero it may or may not have self-inductances, as M = 0 can be satisfied either by setting k = 0 or L = 0.
Inductors in series and parallel

If two coils of inductance L1 and L2 are connected in series, then the effective inductance of the combination of coils
is L = L1 + L2 + 2M … (1)
where M is the coefficient of mutual inductance between the coils.
If the two coils are separated by a large distance
M  0 and Ls = L1 + L2 … (2)
If the two coils are connected in parallel and separated by a large distance, the effective inductance of the combination
L is given by the relation
1 1 1 L1L2
= + … (3) or Lp = … (4)
LP L1 L2 L1 + L2
We see that combination of inductances behave as that of resistances, when mutual inductance is negligible.
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Eddy currents
Consider a solid plate of metal which is introduced into a region having
a magnetic field. Suppose a loop is drawn on the plate, a part of which
 
is inside the field. As the plate moves, the magnetic flux through the    
 
area bound by the loop undergoes a change and induces a current.    
    
There may be a number of such loops on the plate and hence currents    
    
are induced on the surface along various paths. Such currents are
referred to as eddy currents. We do not have a definite conducting loop to guide the induced current. The system
itself looks for the loops on the surface along which eddy currents are induced. Because of the eddy currents in the
metal plate, thermal energy is produced in it. This energy comes at the cost of the kinetic energy of the plate and the
plate slows down. This is known as electromagnetic damping. To reduce electromagnetic damping, one can cut slots
in the plate. This reduces the possible paths of the eddy current considerably.
For example, when we move a metal plate out of a magnetic field, the relative motion between the field and the
conductor again induces a current in the conductor. The conduction electrons making up the induced current whirl
about within the plate as if they were caught in an eddy (or whirlpool) of water. This is called the eddy current.
induced emf e
The magnitude of eddy current is i = =
resistance R
d d / dt
But e = − i=−
dt R
The direction of eddy currents is given by Lenz’s law, or Fleming’s right hand rule.
Eddy currents give rise to loss of energy in devices like transformers, generators and motors. However, eddy currents
are used in induction heaters, speedometer, electromagnetic damping of oscillations etc.
Illustrations
1. A rectangular coil of metallic wire is placed on a uniform field 30 mT with its plane perpendicular to the field. If the
area of loop is shrinking at a constant rate of 0.4 m2s−1, the induced emf in the coil is
(A) 20 mV (B) 18 mV (C) 10 mV (D) 12 mV
Ans (D)
d dA
= − = −B = −(30 10−3 )(0.4) = 12 mV
dt dt
2. An air core solenoid has 1000 turns and is one metre long. Its cross-sectional area is 10 cm2. Its self-inductance is
(A) 2.56 mH (B) 1.92 mH (C) 1.26 mH (D) 2.78 mH
Ans (C)
N
For solenoid, B = 0 ni = 0 i
l
 N  N2A
  = NBA = N  0 i  A =  0 i
 l  l
0 N 2 A −7
(1000 )2
Also  = Li  L = . So, L = 410  10 10−4 = 0.0012566H = 1.26mH
l 1
3. Two coils are wound on the same iron rod so that the flux generated by one, also passes through the other. The first
coil has 100 turns and second has 200 turns. When a current of 2 A flows through the first one, the flux in it is 2.5 
10−4 Wb. The coefficient of mutual inductance of the coil is
(A) 15 mH (B) 25 mH (C) 35 mH (D) 45 mH
Ans (B)
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d di
s = N s s and s = M p
dt dt
d di
 Ns s = M p
dt dt
d  2.5 10−4 − 0 
 M = Ns s = 200  
di p  2−0 
 M = 25 10−3 H = 25 mH
4. When the current through a solenoid increases at a constant rate, the induced current.
(A) is a constant and is in the direction of inducing current.
(B) Is a constant and is opposite to the direction of the inducing current
(C) Increasing with time and is in the direction of inducing current
(D) Increases with time and is opposite to the direction of the inducing current
Ans (B)
 di 
 = L  ;
 dt 
 1  di 
 induced current i = = L  
R R  dt 
di
It is given that is constant. So induced current is constant. From Lenz’s law it opposes the increase in opposing
dt
current.
5. A car moves on a plane road. Induced emf produced across its axle is maximum when it moves
(A) at the poles (B) moves at equator (C) remains stationary (D) no emf is induced at all
Ans (A)
 = Blv sin  . At poles  = 90 and Bv of earth is maximum
  = Blv at poles and is maximum
6. Figure shows a coil placed in a magnetic field decreasing at a rate of 10 Ts −1. There is also a
source of emf 30 V in the coil. The amplitude and direction of the current in the coil are
(A) 2A, anticlockwise (B) 2A, clockwise (C) 4A, anticlockwise (D) 4A, clockwise
Ans (B)
d dB 20
induced = B = A = 2 10 = 20V , iinduced = = 4A , anticlockwise
dt dt 5
30
i = = 6A , clockwise
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i net = 6A − 4A = 2A , clockwise.
7. In the figure space is divided by the line PN into two regions. Region I is field free and
region II has a uniform magnetic field B directed into the plane of the paper. PMN is a
semicircular conducting loop of radius r with centre at ‘O’. The plane of the loop is in the
plane of the paper. The loop is rotated in clockwise direction with a constant angular
velocity  about an axis passing through O and perpendicular to plane of paper. The
resistance of the loop is R. The expression for the magnitude of current in the loop is
Br 2 2Br 2 Br 2 2Br 2
(A)  (B)  (C)  (D) 
2R R 2R R
Ans (A)
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1 r2
dA = (rd)  r = d
2 2
r2
d = BdA = B d
2
d r 2 d Br 2
 = − = −B =− 
dt 2 dt 2
Br 2  Br 2
=  i= = 
2 R 2R
8. A metallic square loop PQRS is moving in its own plane with velocity v in a uniform
magnetic field perpendicular to its plane as shown in the figure. An electric field is
induced
(A) in PS but not in QR (B) in QR but not in PS
(C) neither in PS nor in QR (D) in both PS and QR
Ans (D)
Here, VS − VP = VR − VQ = Blv
[Note: No current flows in the coil because change in magnetic flux linked with the coil is zero]
9. Two magnetic fields exist in the region as shown in figure. A loop abcd of 40 
10 cm2 is placed in the fields. The resistance per unit length of the loop is r = 2
 cm−1. All of a sudden the loop is given a velocity V0 = 20 cms−1 towards right.
What is the potential difference Vc − Vb?
(A) 3 V (B) 2 V (C) −3 V (D) −2V
Ans (C)
1 = Blv = 300  0.1  0.2 = 6 V,
2 = 200  0.1  0.2 = 4V
The equivalent circuit is
 + 6+4
i= 1 2 = = 0.05A
R 200
R cb = 10  2 = 20 
VC + 4 – 0.05  20 = Vb  Vc − Vb = −3V
10. An angle AOB made of a conducting wire moves along its bisector through a magnetic field
B as suggested by figure. The emf induced between the two free ends if the magnetic field is
perpendicular to the plane of the wire is

(A) Blv sin  (B) Blv sin
2

(C) 2Blv sin  (D) 2Blv sin
2
11. A magnet is made to oscillate with a particular frequency, passing through a coil as shown in
the figure. The variation of emf induced across the coil during one cycle with time is best
represented by the graph
(A) (B) (C) (D)
Ans (A)
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When North Pole of the magnet approaches the coil, emf is induced such that left face behaves like North Pole. When
South Pole leaves coil, the emf is induced such that right face behaves like North Pole (thus left face like South Pole).
At the right side external bar is momentarily at rest and thus there has to be a break after half cycle. Similar arguments
hold good when magnet moves from right to left. The correct option is (A).
12. The magnetic field in the cylindrical region, shown in the figure, increases at a constant rate of 20
mTs−1. The two square loops pqrtp and ptuvp have sides each equal to 1 cm and each side has
resistance of 4 . If the switch S1 is closed and S2 is open, the current in the wire pt is
(A) 0.125 A from p and t (B) 0.125 A from t to p
(C) 0.25 A from p to t (D) 0.25 A from t to p
Ans (A)
The flux is associated with ptuv loop only (in the given conditions)
d dB (
= =A = 1110−4 ) (20 10−3 ) = 2 10−6 V
dt dt
 2 10−6
i= = = 0.125 A
R 4 4
As B, increases, South Pole nature of loop ptuv is increasing. So induced current should be in anticlockwise direction
to oppose this, and hence current is from p to t.
13. At what rate should the current change to induce an emf of 60 V?
(A) By changing current at the rate of 10 As−1 (B) By changing current at the rate of 7.5 As−1
(C) By changing current at the rate of 5 As−1 (D) By changing current at the rate of 2.5 As−1
Ans (C)
di di  60
 =L  = = = 5As −1
dt dt L 12
14. A coil has an inductance of 53 mH and a resistance of 0.35 . If a 12 V emf is applied across the coil, the energy is
stored in the magnetic field after the current has built up to its maximum value is
(A) 11 J (B) 21 J (C) 31 J (D) 41 J
Ans (C)
 12
The maximum current in the coil is i0 = = = 34.3A
R 0.35
1 1
So, the maximum energy stored is U0 = Li02 = (53 10−3 )(34.3)2
2 2
Or U 0 31J
15. In the circuit shown, the emf  of the cell, its internal resistance r and the inductances L1 and L2
of the superconducting coils are known. The current established in the coil L, after the key K is
closed is
L1 L2
(A) (B)
r(L1 + L2 ) r(L1 + L2 )
(L1 + L 2 ) (L1 + L 2 )
(C) (D)
L1r L2 r
Ans (B)
If current i is drawn from the battery, then

i = = i1 + i 2 … (i)
r
Inductors L1 and L2 are in parallel and so 1 =  2
di di
Or L1 1 = L2 2
dt dt
On integrating we get L1i1 = L 2i 2 … (ii)
13
L  L
From (ii) i 2 = 1 i1 . Using this in equation (i) = i1 + 1 i1
L2 r L2
 L + L2   L 2
i1  1  =  i1 =
 L2  r r(L1 + L 2 )
16. An inductor of inductance 3 H and resistance 6  is connected to the terminals of a battery of emf 12 V and of
negligible internal resistance. The initial rate of increase of current is
(A) 2 As−1 (B) 3 As−1 (C) 3.5 As−1 (D) 4 As−1
Ans (D)
i = i0 (1 − e )
−t
 … (i)
di −t  1 
 = −i 0e   − 
dt  
i0 − t 
= e … (ii)

 12 L 3
i0 = = = 2 A ,  = = = 0.5 s
R 6 R 6
di 2
 = e = 4 As −1
dt 0.5
[ At the beginning, t = 0]
17. Figure shows a metallic square frame of edge l in a vertical plane. A uniform
magnetic field B is horizontal and perpendicular to the plane of square. Two boys
pull the opposite corners of the square to deform it into a rhombus. They start pulling
the corners at t = 0 and displace the corners at a uniform speed u. The induced emf
in the frame at the instant when the angles at these corners reduce to 60 is
1
(A) Blu (B) 2Blu (C) 2Blu (D) Blu
2
Ans (B)
The loop can be divided into two equal parts. The emf across each part will be
 = B ( l sin 30 2 ) u = Blu ;  net =  +  = 2Blu
18. The rectangular wire frame shown in figure has a width l, mass m, resistance R and
a large length. A constant force F starts acting on the frame pushing it into magnetic
field at time, t = 0. The acceleration of the frame, when its speed has increased to v,
is
2(FR − B2l 2 v) (FR − B2l 2 v) (FR − B2l 2 v) (FR − B2l 2 v)
(A) (B) (C) (D)
mR 2mR 2mR mR
Ans (D)
When the left side of the frame is inside the field and when the speed of frame is v, the emf induced is  = Blv and
 Blv
induced current is i = = . The force F (towards right) exerted by magnetic field is given by
R R
Bl v B2 l 2 v
F' = Bil = B  l =
R R
B2 l 2 v
F−
R = FR − B l v
2 2
Fnet = F − F' = ma  a =
m mR
14
19. The earth’s magnetic field (BE) at the equator is horizontal, uniform and points north – south. A conducting square
loop of side l and resistance R is kept in the vertical plane with two of its sides pointing east-west direction. This loop
is moved in east direction with a velocity v. The current induced in the loop is
B lv 2BElv 4BElv
(A) zero (B) E (C) (D)
R R R
Ans (A)
Magnetic flux linked with the coil is not changing. So  = 0.

So, net emf induced is zero. Thus, the current is zero.
20. *The emf induced between M and Q if the potential between P and Q is 100 V. M is mid point of P and Q.
(A) 25 V (B) 50 V (B) 75 V (C) 100 V

Ans (C)
1
For a rotating conductor of an arbitrary shape in uniform magnetic field B , induced emf is ba = BL2
2
1
PQ = B(2l )2 = 2Bl 2 = 100 V
2
1 1
PM = Bl 2 = PQ = 25V
2 4
( P − Q ) − ( P − M ) = 100 − 25
MQ = 75V
21. A rod of length L and resistance r rotates about one end as shown in figure. Its other end
touches a conducting ring of negligible resistance. A resistance R is connected between the
center and periphery. The current in resistance R is
Bl2 Bl2
(A) (B)
R+r 2(R + r)
Bl 2
Bl 2
(C) (D)
2(R + r) R+r
Ans (A)
 Bl2
i= =
R + r 2(R + r)
15
NCERT LINE BY LINE QUESTIONS

1. Direction of current induced in a wire by moving it in a uniform magnetic field is found
using [NCERT Pg. 215]
(a) Newton’s laws (b) Lenz’s law
(c) Ampere’s rule (d) Right hand grip rule
2. A metallic plate is getting heated. It cannot be due to [NCERT Pg, 218]
(a) A direct current passing through plate
(b) An alternating current passing through it
(c) It is placed Static in Space varying magnetic field but does not vary with time
(d) It is placed in time varying magnetic field
3. A rectangular coil expands on pulling from two diagonal edges in a region of magnetic
field and no emf is induced in the coil. This can be because of [NCERT Pg. 230]
(a) Magnetic field is constant
(b) Magnetic field is in the plane of rectangular coil
(c) Magnetic field has a perpendicular component to the plane of coil whose magnitude is
decreasing
(d) There is a uniform magnetic field perpendicular to plane of coil
4. The self-inductance L of a solenoid of length l and area of cross section A, with fixed number of
turns per unit length increases as [NCERT Pg. 223]
(a) l and A increases
(b) l decreases and A increases
(c) Both l and A decreases
(d) l increases and A decreases
5. The mutual inductance of pair of co-axial neighbouring coils [NCERT Pg. 220]
(a) Increases when they are brought nearer
(b) Increases when one of them is rotated about an axis
(c) Is independent of current passing through coils
(d) Both (a) and (c) are correct
6. A square loop of side length L meter lies in x-y plane in a region, where the magnetic field is
given by B = B0 (i + 2j + 3k) T , Bo is positive constant. The magnitude of magnetic flux passing
through square is [NCERT Pg. 207]
(a) 5B0 L2 Wb (b) 3B0 L2 Wb (C) 14B0 L2 Wb (D) B0 L2 Wb
7. A 20 cm long conductor carrying a current of 10 A is kept perpendicular to magnetic field
of 0.6T. The mechanical power required to move conductor with a speed of 1 ms–1 is
[NCERT Pg. 215]
(a) 1.2 W (b) 1.5 W (c) 0.6 W (d) 0.4 W
8. A square loop of edge 20 cm and resistance of 1  is placed vertically in horizontal plane.
16
A uniform magnetic field of 0.5T is set up across the plane in the direction at 45° to the
plane. The magnetic field is decreased to zero in 0.2 s, at a steady rate. Calculate
magnitude of current induced in this time interval. [NCERT Pg. 208]
(a) 20 mA (b) 50 mA (c) 60 mA (d) 70 mA
9. A circular loop with its plane parallel to plane of paper is entering into uniform magnetic
field directed into the plane of paper perpendicularly. The loop is moved at constant speed V.
Then [NCERT Pg. 212]
(a) No. emf will be induced in the coil
(b) Induced emf is constant in magnitude only
(c) Induced emf is varying with time
(d) Induced emf is constant in magnitude as well as in direction
10. A metallic rod of length 20 cm is rotated with, frequency of 50 rev/s with one end pivoted
at the centre and other end at circumference of circular metallic ring of radius 20 cm about
an axis passing through centre and perpendicular to plane of the ring. A constant and uniform
magnetic field 1.5 T parallel to axis is present everywhere. What is emf induced between centre
and periphery of circular ring. [NCERT Pg. 214]
(a) 2.6 V (b) 9.4 V (c) 4.7 V (d) 12.3 V
11. A cycle wheel with 20 metallic spokes each 1 m long is rotated with speed of 60 rad/s in
a plane normal to horizontal component of earth’s magnetic field BH = 0-5 G at a place.
The emf induced between axle and rim of wheel is [NCERT Pg. 215]
(a) 1.5 mV (b) 12.3 mV (c) 3.0 mV (d) 0.75 mV
12. A conducting arm AB of length 30 cm moves on conducting rails held parallel. A uniform
magnetic field 6 = 0.2 T exists perpendicular to planes of rails. Only the conducting arm
has resistance of 0.5  . The arm is pulled out with constant speed of 20 ms–1, how much force is
required parallel to rails to keep it moving at same speed. [NCERT Pg. 216]
(a) 0.14 N (b) 8 N (c) 16 N (d) 0.25 N
13. Which statement regarding eddy currents among the following is incorrect?
[NCERT Pg. 218]
(a) If rectangular slots are made in copper plate, the magnitude of eddy currents will decrease
(b) Dissipation of heat produced is proportional to strength of eddy currents
(c) Dead beat galvanometer has fixed core made of non-magnetic metallic material
(d) Magnetic brakes in train use the application of eddy current
14. Two circular coils one of small radius r and other of larger radius R (r <<R ) are placed
co-axially with centres coinciding. The mutual inductance of the arrangement is
[NCERT Pg. 221]
0 R 2
 0 r 2
 0 rR 2 0 r 2
(a) (b) (c) (d)
2r 2R (r + R) R
15. A long solenoid is of length 1.25 m and 600 turns per unit length. It is connected to a source
which establishes a current of 2A in circuit. Magnetic energy stored in the solenoid coil with
cross-sectional area 0.1 m2is [NCERT Pg. 224]
17
(a) 0.1 J (b) 0.4 J (c) 0-6 J (d) 1.2 J
16. A rectangular coil of 100 turns with area 0.1 m is rotated at 10 revolution per second and paced
2
in a uniform magnetic field of 0.01 T perpendicular to axis of rotation of the coil. The maximum
voltage generated in coil is [NCERT Pg. 226]
(a) 3.14 V (b) 6.28 V (c) 9.42 V (d) 31.4 V
17. Two thin cylindrical pipes of equal internal diameters made of aluminum and plastic are
taken. The pipes are kept vertical. A small cylindrical magnet without touching sides of
wall of pipe is allowed to fall one by one. Then correct observations are [NCERT Pg. 219]
(a) Magnet takes longer time to cross aluminum pipe
(b) Magnet takes longer time to cross plastic pipe
(c) Eddy currents are generated in aluminum pipe but not in plastic
(d) Both (a) and (c) are correct
18. Which of the following statement is wrong? [NCERT Pg. 225]
(a) In ac generator when flux through coil is maximum, emf induced is minimum
(b) Maximum emf is induced when plane of col is parallel to magnetic field
(c) The emf induced changes periodically with time if coil is rotated at uniform rate
(d) The frequency of rotation of armature coil is 60 Hz in India and 50 Hz in USA
19. A pair of adjacent coils has mutual inductance of 1.5 H. If the Current in one coil changes from
0 to 10 A in 0.5 s. the rate of change of flux linkage with other coil is
[NCERT Pg. 219]
(a) 20 V (b) 30 V (c) 4 V (d) 5 V
20. A circular coil is being deformed into a narrow straight wire at regular stretch. Then
[NCERT Pg. 230)
(a) The direction of induced current is clockwise

(b) The direction of induced current is anticlockwise
(c) Magnetic flux through coil increases
(d) The amount of charge flowing in coil depends on time
NCERT BASED PRACTICE QUESTIONS

1. Whenever the magnetic flux linked with an electric circuit changes, an emf is induced in the circuit. This
is called
(a) electromagnetic induction (b) lenz’s law
18
(c) hysteresis loss (d) kirchhoff’s laws`111
2. According to Faraday’s law of electromagnetic induction
(a) electric field is produced by time varying magnetic flux.
(b) magnetic field is produced by time varying electric flux.
(c) magnetic field is associated with a moving charge.
(d) None of these
3. Lenz’s law is a consequence of the law of conservation of
(a) charge (b) mass (c) energy (d) momentum
4. A magnet is moved towards a coil (i) quickly (ii) slowly, then the induced e.m.f. is
(a) larger in case (i) (b) smaller in case (i)
(c) equal in both the cases (d) larger or smaller depending upon the radius of the coil
5. A current carrying infinitely long wire is kept along the diameter of a circular wire loop, without
touching it, the correct statement(s) is(are)
I. The emf induced in the loop is zero if the current is constant.
II. The emf induced in the loop is finite if the current is constant.
III. The emf induced in the loop is zero if the current decreases at a steady rate.
(a) I only (b) II only
(c) I and II (d) I, II and III
6. An induced e.m.f. is produced when a magnet is plunged into a coil. The strength of the induced e.m.f. is
independent of
(a) the strength of the magnet
(b) number of turns of coil
(c) the resistivity of the wire of the coil
(d) speed with which the magnet is moved
7. A coil of insulated wire is connected to a battery. If it is taken to galvanometer, its pointer is deflected,
because
(a) the induced current is produced
(b) the coil acts like a magnet
(c) the number of turns in the coil of the galvanometer are changed
(d) None of these
8. Two different wire loops are concentric and lie in the same plane. The current in the outer loop (I) is
clockwise and increases with time. The induced current in the inner loop
(a) is clockwise (b) is zero (c) is counter clockwise

(d) has a direction that depends on the ratio of the loop radii.
9. Two identical coaxial circular loops carry a current i each circulating in the same direction. If the loops
approach each other, you will observe that the current in
(a) each increases (b) each decreases
(c) each remains the same (d) one increases whereas that in the other decreases
10. In electromagnetic induction, the induced charge is independent of
(a) change of flux (b) time
(c) resistance of the coil (d) None of these
11. A conducting loop is placed in a uniform magnetic field with its plane perpendicular to the field. An
e.m.f. is induced in the loop, if
(a) it is translated (b) it is rotated about its axis
(c) both (a) and (b) (d) it is rotated about its diameter
19
12. Two coils, A and B, are lined such that emf e is induced in B when the current in A is changing at the rate
I. If current i is now made to flow in B, the flux linked with A will be
(a) (/I)i (b)  i I (c) ( I)i (d) i I/ 
13. Whenever the magnetic flux linked with a coil changes, an induced e.m.f.is produced in the circuit. The
e.m.f. lasts
I. for a short time
II. for a long time
III. so long as the change in flux takes place
The correct statement(s) is/are
(a) I and II (b) II and III (c) I and III (d) III only
14. In a coil of resistance 10 W, the induced current developed by changing magnetic flux through it, is
shown in figure as a function of time. The magnitude of change in flux through the coil in weber is
(a) 8 (b) 2 (c) 6 (d) 4

15. Assertion : Figure shows a horizontal solenoid connected to battery and a switch. A copper ring is
placed on a smooth surface, the axis of the ring being horizontal. As the switch is closed, the ring will
move away from the solenoid.
Reason : Induced emf in the ring, e= d/dt .

(a) Assertion is correct, reason is correct; reason is a correct explanation for assertion.
(b) Assertion is correct, reason is correct; reason is not a correct explanation for assertion
(c) Assertion is correct, reason is incorrect
(d) Assertion is incorrect, reason is correct.
16. The north pole of a bar magnet is moved towards a coil along the axis passing through the centre of the
coil and perpendicular to the plane of the coil. The direction of the induced current in the coil when
viewed in the direction of the motion of the magnet is
(a) clockwise (b) anti-clockwise
(c) no current in the coil (d) either clockwise or anti-clockwise
17. If a current increases from zero to one ampere in 0.1 second in a coil of 5 mH, then the magnitude of the
induced e.m.f. will be
(a) 0.005 volt (b) 0.5 volt (c) 0.05 volt (d) 5 volt
18. A coil has 200 turns and area of 70 cm2. The magnetic field perpendicular to the plane of the coil is 0.3
Wb/m2 and take 0.1 sec to rotate through 180º.The value of the induced e.m.f. will be
(a) 8.4 V (b) 84 V (c) 42 V (d) 4.2 V
19. Assertion : An emf can be induced by moving a conductor in a magnetic field.
Reason : An emf can be induced by changing the magnetic field.
20
20. The north pole of a long horizontal bar magnet is being brought closer to a vertical conducting plane
along the perpendicular direction. The direction of the induced current in the conducting plane will be
(a) horizontal (b) vertical (c) clockwise (d) anticlockwise
21. A metal disc of radius 100 cm is rotated at a constant angular speed of 60 rad/s in a plane at right angles
to an external field of magnetic induction 0.05 Wb/m2. The emf induced between the centre and a point
on the rim will be
(a) 3 V (b) 1.5 V (c) 6 V (d) 9 V
22. A rectangular coil of 100 turns and size 0.1 m × 0.05 m is placed perpendicular to a magnetic field of 0.1
T. The induced e.m.f. when the field drops to 0.05 T in 0.05s is
(a) 0.5 V (b) 1.0 V (c) 1.5 V (d) 2.0 V
23. A solenoid has 2000 turns wound over a length of 0.3 m. Its cross-sectional area is 1.2 × 10–3 m2. Around
its central section a coil of 300 turns is wound. If an initial current of 2 A flowing in the solenoid is
reversed in 0.25 s, the emf induced in the coil will be
(a) 2.4 × 10–4 V (b) 2.4 × 10–2 V (c) 4.8 × 10–4 V (d) 4.8 × 10–2 V
24. A straight conductor of length 2m moves at a speed of 20 m/s. When the conductor makes an angle of 30°
with the direction of magnetic field of induction of 0.1 wbm2 then induced emf
(a) 4V (b) 3V (c) 1V (d) 2V
25. A square coil of side 25cm having 1000 turns is rotated with a uniform speed in a magnetic field about an
axis perpendicular to the direction of the field. At an instant t, the emf induced in the coil is e = 200 sin
100t. The magnetic induction is
(a) 0.50 T (b) 0.02 T (c) 0.01 T (d) 0.1 T
26. When current in a coil changes from 5 A to 2 A in 0.1 s, average voltage of 50 V is produced. The self –
inductance of the coil is :
(a) 6 H (b) 0.67 H (c) 3 H (d) 1.67 H
27. When current i passes through an inductor of self inductance L, energy stored in it is 1/2. L i2. This is
stored in the
(a) current (b) voltage (c) magnetic field (d) electric field
28. Two conducting circular loops of radii R1 and R2 are placed in the same plane with their centres
coinciding. If R1>>R2, the mutual inductance M between them will be directly proportional to
(a) R1/R2 (b) R2/R1 (c) R12 / R2 (d) R22 / R1
29. A coil is wound on a frame of rectangular cross-section. If all the linear dimensions of the frame are
increased by a factor 2 and the number of turns per unit length of the coil remains the same, self-
inductance of the coil increases by a factor of
(a) 4 (b) 8 (c) 12 (d) 16
–1
30. A wire of length 1 m is moving at a speed of 2ms perpendicular to its length in a homogeneous
magnetic field of 0.5 T. The ends of the wire are joined to a circuit of resistance 6W. The rate at which
work is being done to keep the wire moving at constant speed is
(a)1/12W (b)1/6W (c)1/3W (d) 1W
31. A rectangular loop is being pulled at a constant speed v, through a region of certain thickness d, in which
a uniform magnetic field B is set up. The graph between position x of the right hand edge of the loop and
the induced emf E will be
21
(a) (b) (c) (d)
32. A six pole generator with fixed field excitation develops an e.m.f. of 100 V when operating at 1500 r.p.m.
At what speed must it rotate to develop 120V?
(a) 1200 r.p.m (b) 1800 r.p.m (c) 1500 r.p.m (d) 400 r.p.m
33. The self inductance associated with a coil is independent of
(a) current (b) time (c) induced voltage (d) resistance of coil
34. The plane in which eddy currents are produced in a conductor is inclined to the plane of the magnetic
field at an angle equal to
(a) 45° (b) 0° (c) 180° (d) 90°
35. Eddy currents are produced when
(a) A metal is kept in varying magnetic field
(b) A metal is kept in the steady magnetic field
(c) A circular coil is placed in a magnetic field
(d) Through a circular coil, current is passed
36. When strength of eddy currents is reduced, as dissipation of electrical energy into heat depends on the
...A... of the strength of electrical energy into heat depends on the...A... of the strength of electric current
heat loss is substantially ...B ... . Here, A and B refer to
(a) cube, increase (b) inverse, increased
(c) inverse, decreased (d) square, reduced
37. Assertion : Figure shows a metallic conductor moving in magnetic field. The induced emf across its
ends is zero.
Reason : The induced emf across the ends of a conductor is given by e = Bvlsin.
38. When the current in a coil changes from 2 amp. to 4 amp. in 0.05 sec., an e.m.f. of 8 volt is induced in the
coil. The coefficient of self inductance of the coil is
(a) 0.1 henry (b) 0.2 henry (c) 0.4 henry (d) 0.8 henry
39. Two coils of self inductances 2 mH and 8 mH are placed so close together that the effective flux in one
coil is completely linked with the other. The mutual inductance between these coils is
(a) 6 mH (b) 4 mH (c) 16 mH (d) 10 mH
40. The coefficient of self inductance of a solenoid is 0.18mH. If a core of soft iron of relative permeability
900 is inserted, then the coefficient of self inductance will become nearly.
(a) 5.4 mH (b) 162 mH (c) 0.006 mH (d) 0.0002 mH
41. Assertion : When number of turns in a coil is doubled, coefficient of self-inductance of the coil
becomes 4 times.
Reason : This is because L  N2.
22
42. A metallic square loop ABCD is moving in its own plane with velocity v in a unifrom magnetic field
perpendicular to its plane as shown in figure. An electric field is induced
(a) in AD, but not in BC

(b) in BC, but not in AD
(c) neither in AD nor in BC
(d) in both AD and BC
43. A circular coil and a bar magnet placed nearby are made to move in the same direction. If the coil covers
a distance of 1 m in 0.5. sec and the magnet a distance of 2 m in 1 sec, the induced e.m.f. produced in the
coil is
(a) zero (b) 0.5 V (c) 1 V (d) 2 V.
44. Two solenoids of same cross-sectional area have their lengths and number of turns in ratio of 1 : 2 both.
The ratio of self-inductance of two solenoids is
(a) 1 : 1 (b) 1 : 2 (c) 2 : 1 (d) 1 : 4
increased by a factor 2 and the number of turns per unit length of the coil remains the same, self-
inductance of the coil increases by a factor of
(a) 4 (b) 8 (c) 12 (d) 16
46. A square frame of side 10 cm and a long straight wire carrying current 1 A are in the plate of the paper.
Starting from close to the wire, the frame moves towards the right with a constant speed of 10 ms–1 (see
figure). The e.m.f induced at the time the left arm of the frame is at x = 10 cm from the wire is
(a) 2 V (b) 1 V (c) 0.75V (d) 0.5 V

47. A generator has an e.m.f. of 440 Volt and internal resistance of 4000 hm. Its terminals are connected to a
load of 4000 ohm. The voltage across the load is
(a) 220 volt (b) 440 volt (c) 200 volt (d) 400 volt
48. A generator of 220 V having internal resistance r = 10W and external resistance R = 100W. What is the
power developed in the external circuit?
(a) 484 W (b) 400 W (c) 441 W (d) 369 W
49. In the given figure MNPQ which falls through the magnetic field has conductivity s and mass density .
The frame’s terminal velocity assuming it to be small enough so that it reaches its final velocity before
leaving the region occupied by the magnetic field is
Ans-b
23
50. A conducting ring of radius l m kept in a uniform magnetic field B of 0.01 T, rotates uniformly with an
angular velocity 100 rad s–1 with its axis of rotation perpendicular to B. The maximum induced emf in it
is
(a) 1.5V (b) V (c) 2V (d) 0.5V
TOPIC WISE PRACTICE QUESTIONS

Topic 1: Magnetic Flux, Faraday's and Lenz's Law
1. An induced e.m.f. is produced when a magnet is plunged into a coil. The strength of the induced e.m.f. is
independent of
(a) the strength of the magnet
(b) number of turns of coil
(c) the resistivity of the wire of the coil
(d) speed with which the magnet is moved
2. A cylindrical bar magnet is kept along the axis of a circular coil. On rotating the magnet about its axis, the
coil will have induced in it
(a) a current
(b) no current
(c) only an e.m.f.
(d) both an e.m.f. and a current
3. Two identical coaxial coils P and Q carrying equal amount of current in the same direction are brought
nearer. The current in
(a) P increases while in Q decreases
(b) Q increases while in P decreases
(c) both P and Q increases
(d) both P and Q decreases
24
4. A rectangular coil of 20 turns and area of cross-section 25 sq. cm has a resistance of 100  . If a magnetic
field which is perpendicular to the plane of coil changes at a rate of 1000 tesla per second, the current in
the coil is
(a) 1 A (b) 50 A (c) 0.5 A (d) 5 A
5. If a current increases from zero to one ampere in 0.1 second in a coil of 5 mH, then the magnitude of the
induced e.m.f. will be
(a) 0.005 volt (b) 0.5 volt (c) 0.05 volt (d) 5 volt
6. A coil of insulated wire is connected to a battery. If it is taken to galvanometer, its pointer is deflected,
because
(a) the induced current is produced
(b) the coil acts like a magnet
(c) the number of turns in the coil of the galvanometer are changed
(d) None of these
7. The current i in an inductance coil varies with time t according to the graph shown in fig. Which one of the
following plots shows the variation of voltage in the coil with time?
(a) (b) (c) (d)

8. The magnetic flux (in weber) linked with a coil of resistance 10  is varying with respect to time t as  =
4t2 + 2t + 1. Then the current in the coil at time t = 1 second is
(a) 0.5 A (b) 2 A (c) 1.5 A (d) 1 A
9. A coil is suspended in a uniform magnetic field, with the plane of the coil parallel to the magnetic lines of
force. When a current is passed through the coil it starts oscillating; It is very difficult to stop. But if an
aluminium plate is placed near to the coil, it stops. This is due to :
(a) development of air current when the plate is placed
(b) induction of electrical charge on the plate
(c) shielding of magnetic lines of force as aluminium is a paramagnetic material.
(d) electromagnetic induction in the aluminium plate giving rise to electromagnetic damping.
10. A 100 turns coil of area of cross section 200 cm2 having 2  resistance is held perpendicular to a magnetic
field of 0.1 T. If it is removed from the magnetic field in one second, the induced charge produced in it is
25
(a) 0.2 C (b) 2 C (c) 0.1 C (d) 1 C
11. Two identical circular loops of metal wire are lying on a table without touching each other. Loop–A carries
a current which increases with time. In response, the loop–B
(a) remains stationary
(b) is attracted by the loop-A
(c) is repelled by the loop-A
(d) rotates about its CM, with CM fixed (CM is the centre of mass)
12. Consider the situation shown in figure. If the switch is closed and after some time it is opened again, the
closed loop will show
(a) a clockwise current

(b) an anticlockwise current
(c) an anticlockwise current and then clockwise
(d) a clockwise current and then an anti-clock wise current.
13. A magnetic field of 2 × 10–2 T acts at right angles to a coil of area 100 cm2, with 50 turns. The average
e.m.f. induced in the coil is 0.1 V, when it is removed from the field in t sec. The value of t is
(a) 10 s (b) 0.1 s (c) 0.01 s (d) 1 s
14. A rectangular coil of 100 turns and size 0.1 m × 0.05 m is placed perpendicular to a magnetic field of 0.1
T. The induced e.m.f. when the field drops to 0.05 T in 0.05s is
(a) 0.5 V (b) 1.0 V (c) 1.5 V (d) 2.0 V
15. The inductance of a closed-packed coil of 400 turns is 8 mH. A current of 5 mA is passed through it. The
magnetic flux through each turn of the coil is
1 1 1
(a)  0 wb (b)  0 wb (c)  0 wb (d) 0.40 wb
4 2 3
16. The magnetic flux through a circuit of resistance R changes by an amount  in a time t . Then the total
quantity of electric charge Q that passes any point in the circuit during the time t is represented by
 1   
(a) Q = R. (b) Q = . (c) Q = (d) Q =
t R t R t
17. Which of the following figure correctly depicts the Lenz’s law. The arrows show the movement of the
labelled pole of a bar magnet into a closed circular loop and the arrows on the circle show the direction of
the induced current
(a) (b) (c) (d)
26
18. A coil having an area A0 is placed in a magnetic field which changes from B0 to 4B0 in time interval t. The
e.m.f. induced in the coil will be
(a) 3A0B0 / t (b) 4A0B0 / t (c) 3B0 / A0t (d) 4A0 / B0t
19. A horizontal telegraph wire 0.5 km long running east and west in a part of a circuit whose resistance is 2.5
 . The wire falls to g = 10.0 m/s2 and B = 2 × 10–5 weber/ m2, then the current induced in the circuit is
(a) 0.7 amp (b) 0.04 amp (c) 0.02 amp (d) 0.01 amp
20. A coil having n turns and resistance R  is connected with a galvanometer of resistance 4R  . This
combination is moved in time t seconds from a magnetic field W1 weber to W2 weber. The induced current
in the circuit is
(a) −
( W1 − W2 ) (b) −
n ( W2 − W1 )
(c) −
( W2 − W1 ) (d) −
n ( W2 − W1 )
Rnt 5 Rt 5 Rnt Rt
Topic 2: Motional and Static EMI

21. Whenever, current is changed in a coil, an induced e.m.f. is produced in the same coil. This property of the
coil is due to
(a) mutual induction (b) self-induction (c) eddy currents (d) hysteresis
22. The self-inductance of a long solenoid cannot be increased by
(a) increasing its area of cross section
(b) increasing its length
(c) changing the medium with greater permeability
(d) increasing the current through it
23. A metal conductor of length 1 m rotates vertically about one of its ends at angular velocity 5 radians per
second. If the horizontal component of earth’s magnetic field is 0.2×10–4T, then the e.m.f. developed
between the two ends of the conductor is
(a) 5 mV (b) 50 mV (c) 5 mV (d) 50 mV
24. A straight conductor of length 2m moves at a speed of 20 m/s. When the conductor makes an angle of 30°
with the direction of magnetic field of induction of 0.1 wbm2 then induced emf
(a) 4V (b) 3V (c) 1V (d) 2V
25. Two coils have a matual inductance 0.005 H. The current changes in the first coil according to equation I
= I0 sin  t, where I0 = 10A and  = 100  radian/sec. The maximum value of e.m.f. in the second coil is
(a) 2  (b) 5  (c)  (d) 4 
26. A varying current in a coil changes from 10A to zero in 0.5 sec. If the average e.m.f induced in the coil is
220V, the self-inductance of the coil is
(a) 5 H (b) 6 H (c) 11 H (d) 12 H
27. When the current in a coil changes from 8 amp to 2 amp in 3×10–2 seconds, the emf induced in the coil is
2 volt. The self-inductance of the coil is
27
(a) 10 mH (b) 20 mH (c) 5 mH (d) 1 mH
28. A coil of N = 100 tuns carries a current I = 5A and creates a magnetic flux  = 10–5 Tm2 per turn. The value
of its inductance L will be
(a) 0.05 mH (b) 0.10 mH (c) 0.15 mH (d) 0.20 mH
29. In an induction coil the current increases from 0 to 6 amp in 0.3 sec by which induced emf of 30 volt is
produced in it then the value of coefficient of self-inductance of coil will be
(a) 3 henry (b) 2 henry (c) 1 henry (d) 1.5 henry
30. The mutual inductance of a pair of coils is 0.75 H. If current in the primary coil changes from 0.5 A to zero
in 0.01 s, find average induced e.m.f. in secondary coil.
(a) 25.5 V (b) 12.5 V (c) 22.5 V (d) 37.5 V
31. The coefficient of self-inductance of a solenoid is 0.18 mH. If a core of soft iron of relative permeability
900 is inserted, then the coefficient of self-inductance will become nearly.
(a) 5.4 mH (b) 162 mH (c) 0.006 mH (d) 0.0002 mH
32. Two coils are placed close to each other. The mutual inductance of the pair of coils depends upon
(a) relative position and orientation of the two coils
(b) the materials of the wires of the coils
(c) the currents in the two coils
(d) the rates at which currents are changing in the two coils
33. A current of 2.5 A flows through a coil of inductance 5 H. The magnetic flux linked with the coil is
(a) 2 Wb (b) 0.5 Wb (c) 12.5 Wb (d) Zero
34. Two neighbouring coils A and B have a mutual inductance of 20mH. The current flowing through A is
given by i = 3t2 – 4t + 6. The induced emf at t = 2s is
(a) 160 mV (b) 200 mV (c) 260 mV (d) 300 mV
35. When the current in a coil changes from 2 amp. to 4 amp. in 0.05 sec., an e.m.f. of 8 volt is induced in the
coil. The coefficient of self-inductance of the coil is
(a) 0.1 henry (b) 0.2 henry (c) 0.4 henry (d) 0.8 henry
increased by a factor x and the number of turns per unit length of the coil remains the same, self-inductance
of the coil increases by a factor of
(a) x2 (b) x3 (c) x4 (d) x5

37. Two coaxial solenoids are made by winding thin insulated wire over a pipe of cross-sectional area A = 10
cm2 and length = 20 cm. If one of the solenoid has 300 turns and the other 400 turns, their mutual inductance
is (  0 = 4  × 10–7 Tm A–1)
(a) 2.4  × 10–5H (b) 4.8  × 10–4H (c) 4.8  × 10–5 H (d) 2.4  × 10–4 H
28
38. A copper rod of length l is rotated about one end perpendicular to the magnetic field B with constant angular
velocity  . The induced e.m.f. between the two ends is
1 3
(a) Bl 2 (b) Bl 2 (c) Bl 2 (d) 2Bl 2
2 4
39. A conductor of length 0.4 m is moving with a speed of 7 m/s perpendicular to a magnetic field of intensity
0.9 Wb/m2. The induced e.m.f. across the conductor is
(a) 1.26 V (b) 2.52 V (c) 5.04 V (d) 25.2 V
40. A wire of length 1 m is moving at a speed of 2ms–1perpendicular to its length in a homogeneous magnetic
field of 0.5 T. The ends of the wire are joined to a circuit of resistance 6  . The rate at which work is being
done to keep the wire moving at constant speed is
1 1 1
(a) W (b) W (c) W (d) 1W
12 6 3
41. Two identical induction coils each of inductance L are jointed in series are placed very close to each other
such that the winding direction of one is exactly opposite to that of the other, what is the net inductance?
(a) L2 (b) 2 L (c) L /2 (d) zero
42. ( )
A wire of length 1m is perpendicular to x-y plane. It is moved with velocity v = 3iˆ + 3jˆ + 2kˆ m/ s through
( )
a region of uniform induction B = ˆi + 2ˆj T . The potential difference between the ends of the wire is
(a) 1V (b) 1.5V (c) 2.5V (d) 3V

43. A rectangular coil of single turn, having area A, rotates in a uniform magnetic field B with an angular
velocity  about an axis perpendicular to the field. If initially the plane of the coil is perpendicular to the
field, then the average induced emf when it has rotated through 90° is
BA BA BA 2BA
(a) (b) (c) (d)
 2 4 
44. The two rails of a railway track, insulated from each other and the ground, are connected to millivoltmeter.
What is the reading of the millivoltmeter when a train passes at a speed of 180 km/hr along the track, given
that the vertical component of earth’s magnetic field is 0.2 × 10–4 wb/m2 and rails are separated by 1 metre
(a) 10–2 volt (b) 10 mV (c) 1 volt (d) 1 mV
45. If we drop a piece of metal and a piece of non-metl from the same height near the surface of the earth,
which will reach the ground first?
(a) metal
(b) non-metal
(c) both will reach simultaneously
(d) None of these
46. The mutual inductance of a pair of coils, each of N turns, is M henry. If a current of I ampere in one of the
coils is brought to zero in t second, the emf induced per turn in the other coil, in volt, will be
29
MI NMI MN MI
(a) (b) (c) (d)
t t It Nt
47. A coil has 200 turns and area of 70 cm2. The magnetic field perpendicular to the plane of the coil is 0.3
Wb/m2 and take 0.1 sec to rotate through 180º.The value of the induced e.m.f. will be
(a) 8.4 V (b) 84 V (c) 42 V (d) 4.2 V
48. A car moves on a plane road. The induced emf in the axle connecting the two wheels is maximum when it
moves
(a) eastward at equator
(b) westward at equator
(c) eastward at latitude of 45°
(d) at poles
49. A small square loop of wire of side is placed inside a large square loop of side L (L >> ). The loop are
coplanar and their centres coincide. The mutual inductance of the system is proportional is
2
L L2
(a) (b) (c) (d)
L L
50. Two coils of inductances L1 and L2 are linked such that their mutual inductance is M. Then
(a) M = L1 + L2
1
(b) M = ( L1 + L 2 )
2
(c) the maximum value of M is (L1 + L2)
(d) the minimum value of M is L1L2
51. Two coils, one primary of 500 turns and one secondary of 25 turns, are wound on an iron ring of mean
diameter 20 cm and cross-sectional area 12 cm2. If the permeability of iron is 800, the mutual inductance
is :
(a) 0.48 H (b) 2.4 H (c) 0.12 H (d) 0.24 H
52. Two conducting circular loops of radii R1 and R2 are placed in the same plane with their centres coinciding.
If R1>>R2, the mutual inductance M between them will be directly proportional to
(a) R1/R2 (b) R2/R1 (c) R 12 / R 2 (d) R 22 / R1
53. A long solenoid has 500 turns. When a current of 2 ampere is passed through it, the resulting magnetic flux
linked with each turn of the solenoid is 4 ×10–3 Wb. The self- inductance of the solenoid is
(a) 2.5 henry (b) 2.0 henry (c) 1.0 henry (d) 40 henry
54. A metal disc of radius 100 cm is rotated at a constant angular speed of 60 rad/s in a plane at right angles to
an external field of magnetic induction 0.05 Wb/m2. The emf induced between the centre and a point on
the rim will be
(a) 3 V (b) 1.5 V (c) 6 V (d) 9 V
30
55. A copper disc of radius 0.1 m rotated about its centre with 10 revolutions per second in a uniform magnetic
field of 0.1 tesla with its plane perpendicular to the field. The e.m.f. induced across the radius of disc is
 2
(a) volt (b) volt (c)   10 −2 volt (d) 2  10 −2 volt
10 10
56. The current in a coil of L = 40 mH is to be increased uniformly from 1A to 11A in 4 milli sec. The induced
e.m.f. will be
(a) 100 V (b) 0.4 V (c) 440 V (d) 40 V
57. A wire loop is rotated in a uniform magnetic field about an axis perpendicular to the field. The direction of
the current induced in the loop reverses once each
(a) quarter revolution (b) half revolution (c) full revolution (d) two revolutions
Topic 3: Applications of EMI
58. A dynamo converts
(a) mechanical energy into thermal energy
(b) electrical energy into thermal energy
(c) thermal energy into electrical energy
(d) mechanical energy into electrical energy
59. When the speed of d.c. motor increases the armature current
(a) increases
(b) decreases
(c) does not change
(d) increses and decreases continuously
60. If a coil made of conducting wires is rooted between poles pieces of the permanent magnet. The motion
will generate a current and this device is called
(a) electric motor
(b) electric generator
(c) electromagnet
(d) All of the above
61. The armature of a dc motor has 20W resistance. It draws a current of 1.5 A when run by a 220 V dc supply.
The value of the back emf induced in it is
(a) 150 V (b) 170 V (c) 180 V (d) 190 V
62. When a metallic plate swings between the poles of a magnet
(a) no effect on the plate
(b) eddy currents are set up inside the plate and the direction of the current is along the motion of the plate
(c) eddy currents are set up inside the plate and the direction of the current opposes the motion of the plate
(d) eddy currents are set up inside the plate
31
63. A generator has an e.m.f. of 440 Volt and internal resistance of 4000 hm. Its terminals are connected to a
load of 4000 ohm. The voltage across the load is
(a) 220 volt (b) 440 volt (c) 200 volt (d) 400 volt
64. An AC generator of 220V having internal resistance r = 10  and external resistance R = 100  . What is
the power developed in the external circuit?
(a) 484 W (b) 400 W (c) 441 W (d) 369 W
65. A six pole generator with fixed field excitation develops an e.m.f. of 100 V when operating at 1500 r.p.m.
At what speed must it rotate to develop 120V?
(a) 1200 r.p.m (b) 1800 r.p.m (c) 1500 r.p.m (d) 400 r.p.m
66. The number of turns in the coil of an AC generator is 5000 and the area of the coil is 0.25 m2, the coil is
rotated at the rate of 100 turns per second in a magnetic field of 0.2 Weber 1 m2. The peak value of the emf
generated is nearly
(a) 786 KV (b) 440 KV (c) 220 KV (d) 1571 KV
67. Induction furnace is based on the heating effect of
(a) electric field (b) eddy current (c) magnetic field (d) gravitational field
68. The plane in which eddy currents are produced in a conductor is inclined to the plane of the magnetic field
at an angle equal to
(a) 45° (b) 0° (c) 180° (d) 90°
69. The back e.m.f. in a d.c. motor is maximum, when
(a) the motor has picked up max speed
(b) the motor has just started moving
(c) the speed of motor is still on the increase
(d) the motor has just been switched off
70. A series would dc motor has a total resistance of 1.5 ohm. When connected across a 115 volt and running
at a certain speed it draws a current of 10 A. The back emf in the motor is
(a) 100 V (b) 115 V (c) 15 V (d) 1.5 V
NEET PREVIOUS YEARS QUESTIONS

1. The magnetic potential energy stored in a certain inductor is 25 mJ, when the current in the inductor is 60
mA. This inductor is of inductance [2018]
(a) 0.138 H (b) 138.88 H (c) 13.89 H (d) 1.389 H
4
2. A long solenoid of diameter 0.1 m has 2 × 10 turns per meter. At the centre of the solenoid, a coil of 100
turns and radius 0.01 m is placed with its axis coinciding with the solenoid axis. The current in the solenoid
reduces at a constant rate to 0A from 4 A in 0.05 s. If the resistance of the coil is10 2  , the total charge
flowing through the coil during this time is :- [2017]
(a) 16  C (b) 32  C (c) 16   C (d) 32   C
32
3. A long solenoid has 1000 turns. When a current of 4A flows through it, the magnetic flux linked with each
turn of the solenoid is 4 × 10–3 Wb. The self-inductance of the solenoid is : [2016]
(a) 4H (b) 3H (c) 2H (d) 1H
4. An electron moves on a straight line path XY as shown. The abcd is a coil adjacent to the path of electron.
What will be the direction of current if any, induced in the coil? [2015]
(a) adcb
(b) The current will reverse its direction as the electron goes past the coil
(c) No current induced
(d) abcd
5. A conducting square frame of side ‘a’ and a long straight wire carrying current I are located in the same
plane as shown in the figure. The frame moves to the right with a constant velocity ‘V’. The emf induced
in the frame will be proportional to [2015]
1 1 1 1
(a) (b) (c) (d)
( 2x − a )
2
( 2x + a )
2
( 2x − a )( 2x + a ) x2
6. A thin semi-circular conducting ring (PQR) of radius ‘r’ is falling with its plane vertical in a horizontal
magnetic field B, as shown in figure. The potential difference developed across the ring when its speed is
v, is : [2014]
(a) Zero (b) Bv  r2 /2 and P is at higher potential

(c)  rBv and R is at higher potential (d) 2rBv and R is at higher potential
7. A 800 turn coil of effective area 0.05 m is kept perpendicular to a magnetic field 5 × 10–5 T. When the
2
plane of the coil is rotated by 90° around any of its coplanar axis in 0.1 s, the emf induced in the coil will
be : [NEET – 2019]
(1) 2 V (2) 0.2 V (3) 2 × 10–3 V (4) 0.02 V
8. In which of the following devices, the eddy current effect is not used ? [NEET – 2019]
(1) induction furnace (2) magnetic braking in train
(3) electromagnet (4) electric heater
9. A cycle wheel of radius 0.5 m is rotated with constant angular velocity of 10 rad/s in a region of magnetic
field of 0.1 T which is perpendicular to the plane of the wheel. The EMF generated between its centre and
the rim is, [NEET – 2019 (ODISSA)]
33
(1) 0.25 V (2) 0.125 V (3) 0.5 V (4) zero
10. The magnetic flux linked with a coil (in Wb) is given by the equation  = 5t2 + 3t + 16
The magnitude of induced emf in the coil at the fourth second will be [NEET-2020 (Covid-19)]
(1) 33 V (2) 43 V (3) 108 V (4) 10 V
11. A light bulb and an inductor coil are connected to an ac source through a key as shown in the figure
below. The key is closed and after sometime an iron rod is inserted into the interior of the inductor. The
glow of the light bulb [NEET-2020 (Covid-19)]
(1) decreases (2) remains unchanged

(3) will fluctuate (4) increases
12. A wheel with 20 metallic spokes each 1 m long is rotated with a speed of 120 rpm in a plane perpendicular
to a magnetic field of 0.4 G. The induced emf between the axle and rim of the wheel will be, (1 G = 10 –4
T) [NEET-2020 (Covid-19)]
(1) 2.51 × 10–4 V (2) 2.51 × 10–5V (3) 4.0 × 10–5 V (4) 2.51 V
13. Two conducting circular loops of radii R1 and R2 are placed in the same plane with their centres
coinciding. If R1 >> R2, the mutual inductance M between them will be directly proportional to
[NEET-2021]
2 2
R R R R
1) 2 2) 1 3) 2 4) 1
R1 R2 R1 R2
14. Two point charges, q and q are placed at a distance of L as shown in the figure. [NEET-2022]
The magnitude of electric field intensity at a distance R (R>>L) varies as :

1 1 1 1
(1) (2) (3) (4)
R2 R3 R 4 R6
15.. A big circular coil of 1000 turns and average radius 10 m is rotating about its horizontal diameter at 2 rad
s-1 If the vertical component of earth's magnetic field at that place is 2 10−5T and electrical resistance
of the coil is 12.56  then the maximum induced current in the coil will be : [NEET-2022]
1) 0.25A 2) 1.5A 3) 1A 4) 2A
34
NCERT LINE BY LINE QUESTIONS – ANSWERS
1) b 2) c 3) b 4) a 5) d 6) b 7) a 8) d 9) c 10) b
11) a 12) a 13) b 14) b 15) a 16) b 17) d 18) d 19) b 20) b
NCERT BASED PRACTICE QUESTIONS – ANSWERS

1) a 2) a 3) c 4) a 5) a 6) c 7) a 8) c 9) b 10) b
11) d 12) a 13) d 14) b 15) a 16) b 17) c 18) a 19) b 20) d
21) b 22) a 23) a 24) d 25) c 26) d 27) c 28) d 29) b 30) b
31) b 32) b 33) d 34) d 35) a 36) d 37) a 38) b 39) b 40) b
41) b 42) c 43) a 44) b 45) b 46) b 47) d 48) b 49) b 50) b
TOPIC WISE PRACTICE QUESTIONS - ANSWERS

1) 3 2) 2 3) 4 4) 3 5) 3 6) 1 7) 3 8) 4 9) 4 10) 3
11) 3 12) 4 13) 2 14) 1 15) 1 16) 3 17) 1 18) 1 19) 3 20) 2
21) 2 22) 4 23) 2 24) 4 25) 2 26) 3 27) 1 28) 4 29) 4 30) 4
31) 2 32) 1 33) 3 34) 1 35) 2 36) 2 37) 4 38) 1 39) 2 40) 2
41) 4 42) 4 43) 4 44) 4 45) 2 46) 1 47) 1 48) 4 49) 2 50) 4
51) 4 52) 4 53) 3 54) 2 55) 3 56) 1 57) 2 58) 4 59) 2 60) 2
61) 4 62) 3 63) 4 64) 2 65) 2 66) 4 67) 2 68) 2 69) 1 70) 1
NEET PREVIOUS YEARS QUESTIONS-ANSWERS

1) 3 2) 2 3) 4 4) 2 5) 3 6) 4 7) 4 8) 4 9) 2
10) 2 11) 1 12) 1 13) 3 14) 2 15) 3
TOPIC WISE PRACTICE QUESTIONS - SOLUTIONS

Nd d
1. (c)  = −    N and  
dt dt
If so the speed of magnet is fast then correspondingly rate of change of flux is fast & e is maximum. It does
not depend on the resistance of coil.
2. (b) A cylindrical bar magnet is kept along the axis of a circular coil. If the magnet is rotated about its axis,
then there is no change in magnetic flux, there is no emf induced in the coil, no current will be induced in the coil
3. (d) When the coils P and Q are brought nearer, the magnetic flux linked with each coil will increase and
the induced current will induces in the direction opposite to original current according to Lenz law and
hence current in both P and Q decreases.
e dt 20  ( 25 10 ) 1000
−4
4. (c) i = = = = 0.5A
R R 100
5. ( )
(c)  = 5 10 (1/ 0.1) = 0.05V
−3
6. (a) When electric current is passed through the coil inside the galvanometer, induction of magnetic field takes place and thus the
coil acts like a magnet which experiences torque due to permanent magnet inside galvanometer and thus the pointer is deflected.
di
7. (c) emf = L
dt
35
Rate of change of current is constant for one period at a positive value and is constant at negative value for the
second time period. Therefore emf is a constant positive value for first half and constant negative value for second
half.
8. (d) Given :  = 4t 2 + 2t + 1wb
d d
 = ( 4t 2 − 2t + 1) = 8t + 2 = 
dt dt
 8t + 2 8t + 2
Induced current, I = = = A
R 10 10
8 1 + 2
At t = 1s, I = A = 1A
10
9. (d) Because of the Lenz's law of conservation of energy.
10. (c) Initial magnetic flux linked with the coil is
i = BA cos  = 0.1 200 10−4  cos 00 = 2 10 −3 Wb
Final magnetic flux linked with the coil is f = 0
N − N ( f − i )
 By Faraday's law,  = − =
t t
−100 ( 0 − 2 10−3 )
= = 2 10−1 V = 0.2V
1
 0.2V
Induced current I = = = 0.1A
R 2
11. (c) If the current increases with time in loop A, then magnetic flux in B will increase. According to Lenz's
law, loop -B is repelled by loop -A because current in loop B will be antiparallel to that in A.
12. (d) According to Lenz's law, when switch is closed, the flux in the loop increases out of plane of paper, so
induced current will be clockwise
− ( 2 − 1 ) − ( 0 − NBA ) NBA
13. (b) e = = =
t t t
−2 −2
NBA 50  2 10 10
t= = = 0.1s
e 0.1
d d dB
14. (a) e = = ( NBA ) = NA
dt dt dt
 0.1 − 0.05 
= 100  0.1 0.05   = 0.5V
 0.05 
Li 8 10−3  5 10−3
15. (a) N = Li   = =
N 400

= 10−7 = 0 wb
4
 
16. (c) =  = iR   = ( it ) R = QR  Q =
t R
17. (a) When a north pole of a bar magnet moves towards the coil, the induced current in the coil flows in a
direction such that the coil presents its north pole to the bar magnet as shown in figure (a). Therefore, the
induced current flows in the coil in the anticlockwise direction. When a north pole of a bar magnet moves
away from the coil, the induced current in the coil flows in a direction such that the coil presents its south
pole to the bar magnet as shown in figure (b).
36
Therefore induced current flows in the coil in the clockwise direction.
d dBA dB
18. (a) Induced e.m.f  = = = A0
dt dt dt
 1 d
19. (c) i = =
R R dt
( ) (
Here df = B  A = 2 10−5  0.5 10+3  5 )
dt = time taken by the wire to fall at ground
= ( 2h / g ) = (10 /10 ) = 1sec
1/2 1/2
1  ( 2 10 )  ( 0.5 10  5) 

−5 3
i =   = 0.02amp
2.5  1 

d ( W2 − W1 )
20. (b) = R tot = ( R + 4R )  = 5R
dt t
nd −n ( W2 − W1 )
i= = ( W2 & W1 are magnetic flux)
R tot dt 5Rt
21. (b) When current is changed in a coil the magnetic flux is also changed but this flux is because of its own current
and thus, this property is called self induction.
22. (d) The self inductance of a long solenoid is given by
L = r 0 n 2 Al
Self inductance of a long solenoid is independent of the current flowing through it.
23. (b) = 1m,  = 5 rad / s, B = 0.2 10−4 T
B 0.2 10−4  5 1
= = = 50V
2 2
24. (d) Emf = 4 10  0.1 sin 300
1
= 4 = 2 V
2
di d
25. (b) e = M = 0.005  ( i 0 sin t ) = 0.0005  i cos t
dt dt
emax = 0.005 10 100 = 5  cos t = 1
26. (c) Initial current (I1) = 10 A; Final current (I2) = 0; Time (t)
= 0.5 sec and induced e.m.f. (  ) = 220 V.
dI (I − I ) ( 0 − 10 ) = 20L
Induced e.m.f. (  ) = −L = −L 2 1 = −L
dt t 0.5
220
or L = = 11H
20
27. (a) According to Faraday’s law of electro-magnetic inductions,
dI
e= L 2=L
(8 − 2)  L = 10mH
dt 3 10−2
N 100 10−5
28. (d) N = Li  L = = = 0.20mH
i 5
29. (d) I = 6A, t = 0.3s, E = 30V
dI 30  0.3
E=L L = = 1.5H
dt 6
dI 0.5 − 0
30. (d) Given: M = 0.75H and = = 50A / s
dt 0.01
37
 Average induced e.m.f. in secondary coil
dI
e=M = 0.75  50 = 37.5V
dt
31. (b) L = 0 nI
L 
 2 = ----( n and I are same)
L1  0
 L2 = 1L1 = 900  0.18 = 162mH
32. (a) The coupled flux of two coils system is used to define the mutual inductance between the coils. The mutual
N
inductance between the coils is M 21 = 2 21
I1
So it is defined as the proportionality between the emf generated in coil 2 due to the current flows in coil 1. Thus It
depends on the relative position and orientation of two coils
33. (c) Given: current I = 2.5 A
Inductance, L = 5H
Magnetic flux,  = ?
We know,  = LI  5  2.5Wb = 12.5Wb
di
34. (a) i = 3t 2 − 4t + 6 ; = 6t − 4
dt
di di
At t = 2s, = 8 ; ;e = −M = 160 mV
dt dt
di  ( 4 − 2) 
35. (b)  = M or 8 = M  
dt  0.05 
8  0.05
M = = 0.2 henry
2
36. (b) Self inductance = 0 n 2 AL = 0 n 2 AL = 0n 2 (  b )  L
So, when all linear dimensions ( , b and L) are increased by a factor of x. The new self-inductance increases
by a factor of x
 N N A 410−7  300  400 100 10−4
37. (d) M = 0 1 2 =
0.2
= 2.410−4 H
38. (a) If in time t, the rod turns by an angle  , the area generated by the rotation of rod will be
1 1
= l  l = l 2
2 2
So the flux linked with the area generated by the rotation of rod
1  1 1
 = B  l 2  cos 0 = Bl 2 = Bl 2t
2  2 2
d d  1 2  1 2
and so e = =  Bl t  = Bl 
dt dt  2  2
39. (b) Length of conductor (l) = 0.4 m; Speed (v) = 7 m/s and magnetic field (B) = 0.9 Wb/ m2. Induced e.m.f.
(V) = Blv sin  = 0.9 × 0.4 × 7 × sin 90º = 2.52 V.
W  Bvl 
40. (b) Rate of work = = P = Fv; also F = Bil = B  l
t  R 
B 2v 2l 2 ( 0.5)  ( 2 )  (1)
2 2 2
1
P= = = W
R 6 6
38
41. (d) When two inductance coil are joined in series, such that the winding of one is exactly opposite to each
other the emf produced in the two coils are out of phase such that they cancel out.
3 3 2
42. (d) e =  Bvl  ;e=1 2 0
0 0 1
(as length of conductor is 0iˆ + 0 ˆj + 1kˆ )
43. (d) Initially flux,  = BA cos 0 = BA
After rotating through an angle 90°.
Flux through the coil is zero.
So,  = BA
2
Angular speed = w, so, time period = =T

T / 4 is time taken to rotate 90°.
 BA 2 BA
So, = =
t T / 4 
44. ( −4
)
(d)  = Blv = 0.2 10 (1)(180  5 / 8 ) = 10 V = 1mV
−3
45. (b) Non-metallic piece will reach the ground first because there will be no induced current in it due to
movement
d dI NMI
46. (a) E = ( NMI )  E = NM E=
dt dt t
E MI
emf induced per unit turn = =
N t
47. (a) Change in flux = 2 B A N
2  0.3  200  70 10−4
 Induced e.m.f. =
0.1
48. (d) Induced emf in the axle =Blv
v− velocity of car
l− length of car
B− component of magnetic field perpendicular to both l and v .
That is B is the vertical component of magnetic field.
Vertical component of magnetic field is maximum at the poles.
Therefore emf induced in the axle will be maximum at the poles.
2 2i
B=
49. (b) Field at the center of outer square loop = L along the axis.
2 2i
=B 2
= 2
flux linking the smaller square loop L

 2 2 2
=m =
mutual inductance i L
50. (d) the minimum value of M is L1L2
0 r N1 N 2
51. (d) M = A
2 R1
4 10−4  800  500  25 12 10−4
=
2  0.1
52. (d) Mutual inductance between two coil in the same plane with their centers coinciding is given by
39
0  2 2 R22 N1 N2 
M=   henry.
4  R1 
53. (c) Total number of turns in the solenoid, N = 500
Current, I = 2A.
Magnetic flux linked with each turn = 4 × 10–3 Wb
54. (b) Induced emf produced between the centre and a point on the disc is given by
1
e =  BR 2
2
Putting the values,
W = 60 rad / s, N = 0.05Wb / m2 and R = 100cm = 1m
1
We get e =  60  0.05  (1) = 1.5V
2
2
1 1
55. (c) e.m.f. induced = BR 2 = BR 2 ( 2n )
2 2
1
=  ( 0.1)  ( 0.1)  210 = ( 0.1)  volts
2 2
2
LdI 40  10 (11 − 1)
−3
56. (a) e = = = 100V

dt 4  10−3
57. (b) It is because after every 1/2 revolution the current becomes zero and mode of change in flux changes
thereafter (If before the current becomes zero, the mode of flux change was from left to right then after the
current becomes zero the mode of flux change becomes right to left).
58. (d) A dynamo is a device which converts mechanical energy into electrical energy
59. (b) decreases does not change
60. (b) Electric generator works on the principle of EMI in which coil rotated against the magnetic field b/w
the poles causing flux change and hence induced current.
E−e 220 − e
61. (d)  1.5 =  e = 190V
R 20
62. (c) When a changing magnetic flux is applied to a bulk piece of conducting material then circulating
current is called eddy currents are induced
63. (d) Total resistance of the circuit = 4000 + 400 = 4400 W
V 440
Current flowing i = = = 0.1amp
R 4400
Voltage across load = R i = 4000 × 0.1 = 400 volt.
64. (b) V = 200V; r = 10
R | = 10 + 100 = 110
40
V 220
I= = = 2A
R | 100
P = I 2 R = 4 100 = 400W
65. (b) The e.m.f. induced is directly proportional to rate at which flux is intercepted which in turn varies
directly as the speed of rotation of the generator.
66. (d) E = NBA sin t ;  p = NBA
1 2
N = 5 103 ; B = 2 10−1 T ;  = 2f = 2102 ; A = m
4
1
 p = 5 103  2 10−1   2102
4
= 510 = 5 ( 3.14) 104
4
 p = 157.1kV
67. (b) Though most of the times eddy currents are undesirable but they find some useful applications such as
in inductance furnace. Joule's heat causes the melting of a metal piece placed in a rapidly changing
magnetic field.
68. (b) Direction of eddy currents is given by Lenz’s rule.
69. (a) The back e.m.f. in a motor is induced e.m.f., which is maximum, when speed of rotation of the coil is
maximum.
70. (a) If eb is the back emf in the motor, then
 − eb  115 − e b 
i= or 10 =  
R  1.5 
 eb = 100V
NEET PREVIOUS YEARS QUESTIONS-EXPLANATIONS
1. (c) From question energy stored in inductor, U = 25 × 10–3 J
Current, I = 60 mA
1
Energy stored in inductor U = LI 2
2
25 × 10–3 = =  L  ( 60  10−3 )
1 2
2
25  2 106 10−3
L= = 13.89 H
3600
2. (b) Given, no. of turns N = 100
radius, r = 0.01 m
resistance, R =  2  , n = 2 × 104
As we know,
d  N d
 = −N =−
dt R R dt
N d q N 
I = − =−
R dt t R t
41
 N  d  
q = −     t
 R  t  
'–' ve sign shows that induced emf opposes the change of flux.
 2  i   1 0 nN r 2 i
q =  0 nN r    t =
  t   R R
4 10−7 100  4    ( 0.01)  2 104
2
q =
10 2
q = 32 C
3. (d) Here, number of turns n = 100;
current through the solenoid i = 4A;
flux linked with each turn = 4 × 10–3Wb
 Total flux linked, and total = 1000[4 × 10–3] = 4Wb
total = 4  Li = 4  L = 1H
4. (b) Current will be induced,
when e– comes closer the induced current will be anticlockwise
when e– comes farther induced current will be clockwise
5. (c) Emf induced in side 1 of frame e1 = BV

1
0 I
B1 =
2 ( x − a / 2 )
Emf induced in side 2 of frame e2 = B0V
0 I
B2 =
2 ( x + a / 2 )
Emf induced in square frame e = BV 1 − B2V
0 I 0 I
= v− v
2 ( x − a / 2 ) 2 ( x + a / 2 )
1
or e 
( 2 x − a )( 2 x + a )
6. (d) Rate of decreasing of area of semi-circular ring

dA
= = ( 2r ) V
dt
From Faraday’s law of electromagnetic induction
d dA
e=− = −B = − B ( 2rV )
dt dt
As induced current in ring produces magnetic field in upward direction hence R is at higher potential.
7. Given
42
8. Eddy current effect is not used in electric heater

Bl 2 0.1(10 )( 0.5)
2
9. E= ; = = 0.125v
2 2
10.  = 5t 2 + 3t + 60
d
 = = 10t + 3
dt
At t = 4 sec
 = 40 + 3 = 43volt
Impedance, z = R 2 + XL2
11.
X L , Z , I 
1
12.  = Br 2
2
 120   2
 =  ( 0.4 10−4 )   2 
1
  (1)
2   60  
 = 0.810−4 = 2.512 10−4 V
13. Two concentric coils are of radius R1 and R2 as shown
Let current in outer loop be i
43
1 2P 1 2 ( qL )
14. E= . 3 = .
4 0 R 4 0 R3
15. emax = NABW = N  r 2 BW = 1000   r 2  2 10−5  2
emax 1000   (10 )  2 10  2

2 −5
imax = = = 1A
R 12.56
44

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Physics Smart Booklet

Physics Smart Booklet

Faraday’s law of electromagnetic induction

Direction of induced current

through a constant magnetic field.   

Fleming’s right hand rule

Induced electric field

Inductors and inductance

or, Eidt = i R dt + Lidi

Energy density in magnetic field

Relation between mutual inductance and self inductance

Inductors in series and parallel

(A) (B) (C) (D)

Magnetic flux linked with the coil is not changing. So  = 0.

(A) 25 V (B) 50 V (B) 75 V (C) 100 V

NCERT LINE BY LINE QUESTIONS

(a) The direction of induced current is clockwise

NCERT BASED PRACTICE QUESTIONS

(a) is clockwise (b) is zero (c) is counter clockwise

(a) 8 (b) 2 (c) 6 (d) 4

Reason : Induced emf in the ring, e= d/dt .

(a) (b) (c) (d)

(a) in AD, but not in BC

(a) 2 V (b) 1 V (c) 0.75V (d) 0.5 V

TOPIC WISE PRACTICE QUESTIONS

(a) (b) (c) (d)

(a) a clockwise current

(a) (b) (c) (d)

Topic 2: Motional and Static EMI

(a) x2 (b) x3 (c) x4 (d) x5

(a) 1V (b) 1.5V (c) 2.5V (d) 3V

NEET PREVIOUS YEARS QUESTIONS

(a) Zero (b) Bv  r2 /2 and P is at higher potential

(1) decreases (2) remains unchanged

The magnitude of electric field intensity at a distance R (R>>L) varies as :

NCERT BASED PRACTICE QUESTIONS – ANSWERS

TOPIC WISE PRACTICE QUESTIONS - ANSWERS

NEET PREVIOUS YEARS QUESTIONS-ANSWERS

TOPIC WISE PRACTICE QUESTIONS - SOLUTIONS

1  ( 2 10 )  ( 0.5 10  5) 

flux linking the smaller square loop L

56. (a) e = = = 100V

5. (c) Emf induced in side 1 of frame e1 = BV

6. (d) Rate of decreasing of area of semi-circular ring

8. Eddy current effect is not used in electric heater

Let current in outer loop be i

emax 1000   (10 )  2 10  2

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