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 Bihar Board Class 12th Physics Solutions Chapter 6 Electromagnetic Induction

Bihar Board Class 12 Physics Electromagnetic Induction Textbook Questions and Answers

Question 1.
Predict the direction of induced current in the situations described by the following Figs, (a) to (f).



Answer:
(a) As the magnet moves towards the solenoid, the magnetic flux linked with the solenoid increases. According to Lenz’s law, the induced e.m.f. produced in the solenoid is such that it opposes the very cause producing it i.e., it opposes the motion of the magnet. Hence the face q of it becomes the south pole and p becomes north pole. Therefore, the current will flow along pq in the coil i.e., along q r pq in this figure i.e., clockwise when seen from the side of the magnet according to clock rule.

(b) As the north pole moves away from xy coil, so the magnetic flux linked with this coil decreases. Thus according to Lenz’s law, the induced e.m.f. produced in the coil will oppose the motion of the magnet. Hence the face. X becomes S-pole, so the current will flow in the clockwise direction i.e., along yzx in the cone. For coil pq, the sou th pole of the magnet mov es towards end q and thus this end will acquire south polarity so as to oppose the motion of the magnet, hence the current will flow along prq in the coil.

(c) The induced current will be in the anticlockwise direction, i.e., along yzx.
(d) The induced current will be in the clockwise direction i.e., along zyx.
(e) The battery current in the left coil will be from right to left, so by mutual induction, the induced current in the right coil will be in the opposite direction i.e., from left to right or along xry.
(f) In this case, there is no change in magnetic flux linked with the wire, so no current will flow through the wire since there is no induced current as the field lines lie in the plane of the loop.

Question 2.
Use Lenz’s law to determine the direction of induced current in the situation described by Fig. given below:
(a) A wire of irregular shape turning into a circular shape;
(b) A circular loop being deformed into a narrow straight wire.

Answer:
(a) When a wire of irregular shape turns into a circular loop, the magnetic flux linked with the loop increases due to increase in area. The circular loop has greater area than the loop of irregular shape. The induced e.m.f. will cause current to flow in such a direction so that the wire forming the loop is pulled inward from all sides i.e., current must flow in the direction adcl^a as shown in Fig. (a) i.e., in anticlock-wise direction so that the magnetic field produced by the current (directed out of the paper) opposes the applied field.

(b) In Fig. (b), a circular loop deforms into a narrow straight wire i.e., upper side of loop should move downwards and lower end should move upwards to oppose the motion of the circular loop, thus its area decreases as a result of which the magnetic flux linked with it decreases. To oppose the decrease in magnetic flux, the induced current should flow anti clockwise in the loop, i.e., along a’d’c’b’a’. Due to the flow of anti-clockwise current, the magnetic field produced will be out of the
page and hence the applied field is supplemented.

Question 3.
A long solenoid with 15 turns per cm has a small loop of area 2.0 cm2 placed inside the solenoid normal to its axis. If the current carried by the solenoid changes steadily from 2.0 A to 4.0 A in 0.1 s, what is the induced emf in the loop while the current is changing?
Answer:
Here, n = number of turns per unit length of the solenoid = 15 turns cm-1
= 1500 turns m-1.
A = area of loop placed insider the solenoid = 2.0 cm2 = 2 x 10-4 m2.
I1 = 2.0 A, I2 = 4.0 A,
dl = I2 – I1 = 4 – 2 = 2A
dt = 0.1 s, dIdt = 20.1 = 20 As-1,
e = induced e.m.f. in the loop = ?
We know that the magnetic field produced inside the solenoid is given by :
B = µ0 nl
lf ø be the magnetic flux linked with the ioop, then
ø = BA = µ0nlA.
∴ Using the relation, e = – dødt we get
e = –ddt(ø) = –ddt(m0 nlA) = -µ0nAdldt
∴ magnitude of e is given by
e = µ0nA.dldt
= 4n x 10-7 x 1500 x 2 x 10-4 x 20
= 7.54 x 10-6 V = 7.5 x 10-6V.

Question 4.
A rectangular wire loop of sides 8 cm and 2 cm with a small cut is moving out of a region of uniform magnetic field of magnitude 0.3T directed normal to the loop. What is the emf developed across the cut if the velocity of the loop is 1 cm s-1 in a direction normal to the (i) longer side, (ii) shorter side of the loop? For how long does the induced voltage last in each case?

Answer:
Here, l = length of loop = 8 cm, b = breadth of loop = 2 cm
B = magnetic field = 0.3 T
v = Velocity of the loop = 1 cms-1 = 102ms-1
A = Area of loop = l x b = 8 x 2 x 10-4 m2 = 16 x 10-4 m2.
e = induced e.m.f. = ?
t = time for which e.m.f. lasts in the loop = ?

(i) When the velocity is normal to the longer side :
Using the relation, e = B/v, we get
e = 0.3 x 8 x 10-2 x 10-2 = 2.4 x 10-4 V = 0.24 x 10-3 V = 0.24 mV.

The e.m.f. will last in the loop till it does not get out of the magnetic field i.e., for the time the loop takes to travel a distance equal to the length of the shorter arm.

Question 5.
A 1.0 m long metallic rod is rotated with an angular frequency of 400 rad s-1 about an axis normal to the rod passing through its one end. The other end of the rod is in contact with a circular metallic ring. A constant and uniform magnetic field of 0.5 T parallel to the axis exists everywhere. Calculate the emf developed between the centre and the ring.
Answer:
Here, l = length of the metallic rod = 1.0 m.
ω = angular frequency = 400 rads-1.
B = Uniform magnetic field = 0.5 T
υ1 = Velocity of one end = 0.

Let e = induced e.m.f. between the centre and the ring = ?
υ2= Velocity of 2nd end = ωl (∵ υ = rω, here r = l)
If υ be the average velocity, then
υ = v1+v22=0+ωl2=ωl2
e = induced e.m.f. between the centre and the ring = ?
Using the relation, e = Bvl, we get
e = Bωl.12 = 12 Bω l2
= 12 x 0.5 x 400 x (1)2
= 100 V

Question 6.
A circular coil of radius 8.0 cm and 20 turns is rotated about its vertical diameter with an angular speed of 50 rad in a uniform horizontal magnetic field of magnitude 3.0 x 10-2 T. Obtain ,the maximum and average emf induced in the coil. If the coil forms a closed loop of resistance 100, calculate the maximum value of current in the coil. Calculate the average power loss due to Joule heating. Where does this power come from?
Answer:
Here, n = number of turns in the coil = 20
r = radius of coil = 8.0 cm = 8 x 10-2 m
ω = angular speed of the coil = 50 rad s-1
B = magnetic field = 3.0 x 10-2 T.
Let e0 be the maximum e.m.f. in the coil = ?
and eav be the average e.-m.f. in the coil = ?
We know that the instaneous e.m.f. produced in a coil is given by :
e = BA nW sin ωt.
for e to be maximum emax., sin ωt = 1.
emax = BAnω = B. πr2 nω

where A = πr2 is the area of the coil :

eav for one cycle is given by

i.e., eav is zero as the average value of sin wt for one complete cycle is always zero.
Now R = resistañof the closed loop formed by the coil = 10
Let Imax = max maximum current in the coil = ?
∴ Using the relation,

∴ eav for one cycle is given by

The induced current causes a torque opposing the rotation of the coil. An external agent must supply torque and do work to counter this torque in order to keep the coil rotating uniformly. Thus the source of the power dissipated as heat in the coil is the external agent i.e., rotor.

Question 7.
A horizontal straight wire 10 m long extending from east to west is falling with a speed of 5.0 m s-1, at right angles to the horizontal component of the earth’s magnetic field, 0.30 x 10-4 Wb m2.
(a) What is the instantaneous value of the emf induced in the wire?
(b) What is the direction of the emf?
(c) which end of the wire is at the higher electrical potential?
Answer:
Here, l = length of wire = 10 m.
υ = speed = 5.0 ms-1.
BH = horizontal component of earth’s magnetic field = 0.30 x 10-4 Wb m-2.

(a) Let e be the value of instantaneous induced e.m.f. in the wire.
∴ Using the relation,
e = Blυ,we get
e = BH lυ
= 0.3 x 10-4 x 10 x 5
= 1.5 x 10-3 V.
(b) The induced e.m.f. will act in a direction so as to oppose the motion of the falling straight wire i.e., from west to east.
(c) As the induced e.m.f. opposes its cause i.e. it sets from low to
high potential end, thus eastern end is at higher potential as induced e.m.f. acts from West to East.

Question 8.
Current in a circuit falls from 5.0 A to 0.0 A in 0.1 s. If an average emf of 200 V induced, give an estimate of the self-inductance of the circuit.
Answer:
Here, I1 = initial current in the circuit = 5.0 A
I2 = Final current in the circuit = 0.0 A
∴ Change in current, dl = I2 – I1 = 0 – 5 = – 5 A
dt = time in which the current changes = 0.1 s
e = average induced e.m.f. = 200 V
L = Self inductance of the circuit = ?
Using the relation,
e = -L dldt
we get

Question 9.
A pair of adjacent coils has a mutual inductance of 1.5 H. If the current in one coil changes from 0 to 20 A in 0.5 s, what is the change of flux linkage with the other coil?
Answer:
Here, M = Mutual inductance of two coil = 1.5 H
dl = change in current in one coil = I2 – I1 = 20 – 0 = 20 A.
dt = time in which the change in current takes place = 0.5 s.
dø = change in magnetic flux of the other coil = ?
If e be the induced e.m.f. produced in the other coil, then
Using the relation,
e = -M dIdt, we get
e = – 1.5 x 200.5 = – 60 V
Also we know that
e = – dødt
dø = -e x dt = -(-60) x 0.5 = 30 Wb.

Question 10.
A jet plane is travelling towards west at a speed of 1800 km/h. What is the voltage difference developed between the ends of the wing having a span of 25 m, if the Earth’s magnetic field at the location has a magnitude of 5 x 10-4 T and the dip angle is 30°.
Answer:
Here, v→ = 1800 km/h = 1800 x 518= 500 ms-1 towards west.
l = wingspan = 25 m, B = earth’s magnetic field = 5 x 10-4T.
δ = angle of dip = 30°
.’. Bv = Vertical component of earth’s magnetic field
= B sin δ = 5 x 10-4 x sin 30 = 5 x 10-4 x 12 = 2.5 x 10-4 T.
Bv is normal to both the wings and the direction of motion. So if e be the induced e.m.f. produced (= voltage difference developed between the ends of the wing), then
e = Bvl = 2.5 x 10-4 x 25 x 500 = 3125 x 10-3 V.
= 3.125 V = 3.1 V (Using significant figures).
The direction of the wing is immaterial as long as it is horizontal for this answer

Question 11.
Suppose the loop in Exercise 6.4 is stationary but the current feeding the electromagnet that produces the magnetic field is gradually reduced so that the field decreases from its initial value of 0.3 T at the rate of 0.02 T s-1. If the cut is joined and the loop has a resistance of 1.6 Ω, how much power is dissipated by the loop as heat? What is the source of this power?
Answer:
Here, A = 8 x 2 cm2 = 16 x 104 m2.
B1 = initial value of magnetic field = 0.3 T
dBdt = rate of decrease of the magnetic field = 0.02 Ts-1.
R = resistance of the loop = 1.6 Ω
P = power dissipated as heat = ?
‘ If 6 be the flux linked with the loop, then ø = BA
Thus the induced e.m.f. ‘e’ produced in the loop is given by

the induced current I is given by

The source of this power is the external agency changing the magnetic field with time.

Question 12.
A square loop of side 12 cm with its sides parallel to X and Y axes is moved with a velocity of 8 cm s’1 in the positive x-direction in an environment containing a magnetic field in the positive z-direction. The field is neither uniform in space nor constant in time. It has a gradient of 10-3 T cm-1 along the negative x-directipn (that is it increases by 10-3 T cm-1 as one moves in the negative x-direction), and it is decreasing in time at the rate of 10″3 T s-1. Determine the direction and magnitude of the induced current in the loop if its resistance is 4.50 mΩ.
Answer:
Here, a = side of the square loop = 12 cm = 12 x 10-2 m.

v→ = Velocity of loop parallel to x-axis = 8 cms-1 = 8 x 10-2 ms-1
Let B = variable magnetic field acting away from us ⊥ ar to the XY plane along z axis i.e., plane of paper represented by x.
dBdX 10-3 T cm-1 = 10-3 x 102 T m-1 = o.l T m-1 = field gradient along -υe x direction.
dBdX = rate of variation with time
= 10-3 Ts-1
R = resistance of the loop = 4.50 mΩ = 4.5 x 103 Ω.
Let I = induced current = ? and its direction = ?
∴ A = area of loop = a2 = (12 x 10-2)2 m2 = 144 x 10-4m2.
The magnetic flux changes (i) due to the variation of B with time
and (ii) due to motion of the loop in non-uniform B→ .
Thus if ø be the total magnetic flux of the loop, then ø is calculated as :
Area of shaded part = adx

Let dø = magnetic flux linked with shaded part = B (x, t) a dx



Clearly the two effect add up as these cause a decrease in flux along the + z direction.
∴If e be the induced e.m.f. produced, then

The direction of induced current is such as to increase the flux through the loop along + z-direction. Thus if for the observer, the loop moves to the right, the current will be seen to be anti-clockwise.

Question 13.
It is desired to measure the magnitude of field between the poles of a powerful loud speaker magnet. A small flat search coil of area 2 cm2 with 25 closely wound turns, is positioned normal to the field direction, and then quickly snatched out of the field region. Equivalently, one can give it a quick 90° turn to bring its plane parallel to the field direction). The total charge flown in the coil (measured by a ballistic galvanometer connected to coil) is 7.5 mC. The combined resistance of the coil and the galvanometer is 0.50 Q. Estimate the field strength of magnet.
Answer:
Here, A = area of coil = 2 cm2 = 2 x 104 m2.
N = number of turns = 2 s.
q = charge flown in the coil = 7.5 mC = 7.5 x 10-3 C
R = resistance of coil and galvanometer = 0.50 Ω.
Let B = field strength of the magnet = ?
Let ø1 = magnetic flux linked with the search coil initially :
.’. ø1 = NBA.
Also let 2 = magnetic flux linked with the coil finally when it is brought out of the field.
ø2 = 0
If e be the induced e.m.f. produced in the coil, then
e = – dødt
According to Ohm’s Law, e = IR, where I = induced current = dqdt

Question 14.
Figure shows a metal rod PQ resting on the smooth rails AB and positioned between the poles of a permanent magnet. The rails, the rod, and the magnetic field are in three mutual perpendicular directions. A galvanometer G connects the rails through a switch K. Length of the rod = 15 cm, B = 0.50 T, resistance of the closed loop containing the rod = 9.0 mO. Assume the field to be uniform.

(a) Suppose K is open and the rod is moved with a speed of 12 cm s-1 in the direction shown. Give the polarity and magnitude of the induced emf.
(b) Is there an excess charge built up at the ends of the rods when K is open? What if K is closed?
(c) With K open and the rod moving uniformly, there is no net force on the electrons in the rod PQ even though they do experience magnetic force due to the motion of the rod. Explain.
(d) What is the retarding force on the rod when K is closed?
(e) How much power is required (by an external agent) to keep the rod moving at the same speed (=12 cm s-1) when K is closed? How much power is required when K is open?
(f) How much power is dissipated as heat in the closed circuit? What is the source of this power?
(g) What is the induced emf in the moving rod if the magnetic field is parallel to the rails instead of being perpendicular?
Answer:
Here, B = 0.50 T
l = length of the rod = 15 cm = 15 x 10-2 m.
R = resistance of the closed loop containing the rod = 9.0 mΩ = 9 x 10-3Ω.
(a) v = speed of the rod = 12 cms-1 = 12 x 10-2 ms-1.
The magnitude of the induced e.m.f. is
E = Blυ = 0.50 x 15 x 10-2 x 12 x 10-12
= 9 x 10-3V
According to Fleming’s left hand rule, the direction of Lorentz
force F→ = -e( V→ x B→ on electrons in PQ is from P to Q. So the end P of the rod will acquire positive charge and Q will acquire negative charge.

(b) Yes. When the switch K is open, the electrons collect at the end Q, so excess charge is built up at the end Q. But when the switch K is closed, the accumulated charge at the end Q is maintained by the continuous flow of current.

(c) This is because the presence of excess charge at the ends P and Q of the rod sets up an electric field E . The force due to the electric field
(q E→) balances the Lorentz magnetic force q (v→ x B→ ). Hence the net force on the electrons is zero.

(d) When the key K is closed, current flows through the rod. The retarding force experienced by the rod is:
F=BIl = B(ER)
Where I = ER is the induced current.

(e) The power required by the external agent against the above retarding force to keep the rod moving uniformly at speed 12 ems-1 when ‘ K is dosed is given by ,
P = FV = 7.5 x 10-2 x 12 x 10-2
= 90 x 10-3 W = 9 x 10-3 W.

(f) Power dissipated as heat is given by

The source of this power is the power provided by the external agent calculated in (e)

(g) Zero. This is because when the magnetic field is parallel to therails, θ = 0°, so induced e.m.f. E = Blυ sin θ = 81v sin 0 = 0. In this situation, the moving rod does not cut the field lines, so there is no change in the magnetic flux, hence E = 0.

Question 15.
An air-cored solenoid with length 30 cm, area of cross – section 25 cm2 and number of turns 500, carries a current of 2.5 A. The current is suddenly switched off in a brief time of 10-3 s. How much is the averageback emf induced across the ends of the open switch in the circuit? Ignore the variation in magnetic field near the ends of the solenoid.
Answer:
Here, l = length of the solenoid = 30cm = 30 x 10-2m.
A = area of cross-section = 25 cm2 = 25 x 10-4 m2.
N = total no. of turns = 500
I1 = current in the solenoid before switching = 2.5 A
I2 = current in the solenoid after switching = 0
dt = time of change of current 10-3 s.
Let E average back e.m.f. induced in the solenoid.
If n be the no. of turns per unit length, then
n = Nl=50030×10−2=50003
We know that the magnetic field inside the solenoid is given by
B = µ0nI = 4π x 10-7 x 50003. x 2.5 = 52.36 x 10-4T.
If ø1 be the initial magnetic flux linked with the solenoid,
Then ø1 = NBA = 5000 x 52.36 x 10-4 x 25 x 10-4 = 65.45 x 10-4 Wb.
ø2 = final flux = 0 as I2 = 0.

∴ dø = change in magnitude flux = ø2 – ø1
= – 65.45 x 10-4 Wb.
∴ E = dødt = 65.45×10−410−3 = 6.545 V.

Question 16.
(a) Obtain an expression for the mutual inductance between a long straight wire and a square loop of side a as shown in Fig. here.
(b) Now assume that the straight wire carries a current of 50 A and the loop is moved to the right with a constant velocity, v = 10 m/s. Calculate the induced emf in the loop at the instant when x = 0.2 m. Take ∝ = 0.1 m and assume that the loop has a large resistance.

Answer:
(a) Here, the side of the square loop = a
Current in the long straight wire = I
The straight current carrying wire produces magnetic field around it.
Hence a magnetic flux is linked with the square ioop moving with a velocity υ ms-1.
Let B be the magnetic field produced by the long straight wire at a distances r from it, then
B = μ04π⋅2Ir ……(1)
and it acts normal to the plane of the loop.
Let the square ioop be divided into a large number of elementary strips. Consider one such styip of width dr at a distance r from the wire.
If dA be the area of this strip, then
dA = adr.
If dø be the magnetic ílpx linked with the strip, then dø = BdA =
μ04π⋅2Ir = μ02π⋅2Ir ……(2)
If ø be the total magnetic flux linked with the square loop, then

Which is the required expression for M.
(b) Here, x = 0.2 m, a = 0.1 m,
V = 10 ms-1,
I= 50 A.
Let e = induced e.m.f. in the loop = ?
We know that

Putting values of I, a, υ, x, we get

Question 17.
A line charge X per unit length is lodged uniformly onto the rim of a wheel of mass M and radius R. The wheel has light non¬conducting spokes and is free to rotate without friction about its axis as shown in fig here. A uniform magnetic field extends over a circular region within the rim. It is given by,
B = -B0k (r<a;a<R)
= 0 (otherwise)
What is the angular velocity of the wheel after the field is suddenly switched off?

Answer:
Let ω be the angular velocity of the wheel of mass M and radius R.
Let e = Induced e.m.f. produced.
The rotational K.E. of the rotating wheel = 12 Iω2 ……….(1)
where I = Moment of inertia of wheel
= 12 MR2 …………(2)

Also by clef. ole.m.f.,

or Work done = eQ
Applying the work energy theorem, we get
Rotational K.E. = Work done
or Rotational K.E. = Q x e …..(3)
We know that the e.m.f. of a rod rotating in a uniform magnetic
field is given by 12 Bωa2, since here the magnetic field is changing, we assume the average over the time span and thus average value of e.m.f. is given by

Aliter:
When the B→ is suddenly brought to zero, change in magnetic field occurs and is given by dB→dt=−ddt(B0k^) ………..(1)

Due to change in magnetic field, an electric field E is produced
which exerts a force qE on a mobile charge q. The work done in moving
the charge once around the ioop is given by
W = Force distance
= qE.2πR
if e = induced e.m.f., then
e = Wq = qEq.2πR=E.2πR ………(3)
Also we know that

Also we know that impulse due to electric field is given by
J = F. dt = qE dt = change in momentum of wheel
or qE dt = mv – 0 = m. Rw
or E = mRωqdt ……….(6)

where ω = angular speed of wheel
∴ From (5) and (6), we get

Hence, proved.