# NCERT : Electromagnetic Induction

Q: 1. Predict the direction of induced current in situations described by the following Figures (see in Textbook)

Sol.
(a) South pole developes at q, current induced must be clockwise at q.
In the coil, induced current is form p to q.

(b) Coil pq in this case would develop s pole at q and coil XY would also develop S pole at X. Therefore induced current in coil pq
will be form q to p and induced current in coil XY will be form X to Y

(c) Induced current in the right loop will be along XYZ.

(d) Induced current in the left loop will be along ZYX as seen from front.

(e) Induced current in the right coil is from X to Y

(f) No current is induced because magnetic lines of force lie in the plane of the loop.

Q:2. Use Lenz’s law to determine the direction of induced current in the situation described by Fig.(see in Textbook)

(a) A wire of irregular shape turning into a circular shape
(b) a circular loop being deformed into a narrow straight wire. The crosses indicate the magnetic field into the paper and the dots indicate magnetic field out of the paper.

Sol:
(a) When a wire of irregular shape turns into a circular loop, area of the loop trends to increases. Therefore, magnetic. flux linked with the loop increases. According to Lenz’s law the direction of induced current must oppose the magnetic field, for which induced current should flow along adcba

(b) In this case, the magnetic flux tends to decrease. Therefore, induced current must support the magnetic field, for which induced current should flow along adcba.

Q: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 ?

Sol. Number of turns per unit length , n = 15 cm-1 = 1500 m-1
Area , A = 2 cm2 = 2 × 10 m-4 , μo = 4π × 10 -7

Change in current , dI = 4-2 = 2A ,

Change in time , dt = 0.1 sec

Magnetic field B = μo n I

Magnetic Flux , φ = B A = (μo n I) A

A/c to Formula , induced emf , $\displaystyle e = -\frac{d\phi}{dt}$

$\displaystyle e = -\frac{d(\mu_0 nIA)}{dt}$

$\displaystyle e = -(\mu_0 n A)\frac{dI}{dt}$

On putting all the given values we get

e = − 7.54 × 10-6

Q: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 0.3 T 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
(a) longer side,
(b) shorter side of the loop? For how long does the induced voltage last in each case ?

Sol. using formula ,  e = B l v  and time t = length of wire / v

(a) along longer side

length , l = 8 cm = 0.08 m  , B = 0.3 T , v= 1 cm/s = 0.01 m/s

emf developed , e= B l v = 0.3 × 0.08 × 0.01 = 0.24 mV

time of emf = length of shorter arm/v = 0.02/0.01 = 2 sec.

(since , emf developed as long as loop does not get out the field , i.e. distance travelled by shorter arm )

(b)along longer side

emf developed , e= B l v =0.3 x 0.02 x 0.01 = 0.06mV

time of emf = length of longer arm/v = 0.08/0.01 =8 sec.

Q :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.

Sol . Induced emf , e = Blv

Average velocity of the rod , v = (0 + ωl)/2 = ωl/2

Induced emf , $\displaystyle e = \frac{1}{2}B\omega l^2$

On putting the values , B = 0.5 T , l = 1 m ,  ω = 400 rad s–1

e = 100 V .

Q:6. A circular coil of radius 8.0 cm and 20 turns rotates about its vertical diameter with an angular speed of 50 s-1 in a uniform horizontal magnetic field of magnitude 3 × 10-2T. Obtain the maximum and average e.m.f. induced in the coil. If the coil forms a closed loop of resistance. 10 Ω, calculate the maximum value of current in the coil. Calculate the average power loss due to joule heating. Where does this power come from?

Sol.

Here, r = 8.0 cm = 8×10-2 m, N = 20, ω = 50 s-1, B = 3 × 10-2 T, e0 = ?,eav = ?,R = 10 Ω, P = ?

As e0 = NAB ω

= N(π r2 )Bω

e0 = 20 × (22/7) (8×10-2)2 × 3×102×50

= 0.603 volt

Average value of emf induced over a full cycle, eav = 0

l0 = e0/R

= 0.603/10 = 0.0603 A

Power dissipated = (e0l0 )/2

= (0.603×0.0603)/2 = 0.018 W

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

Q:7. A horizontal straight wire 10 m long extending from east to west is falling with a speed of 5.0 ms-1 at
Right angles of the horizontal component of the earth’s magnetic field 0.30 ×10-4 Wb m2.
What is the instantaneous value of the e.m.f induced in the wire ?.
(b) what is the direction of the e.m.f. ? (b) which end of the wire is at higher electric potential

Sol .
Here,l = 10 m, v = 5.0 ms-1 ; B = 0.30 × 10-4 T

e = B l v

= 0.30 × 10-4× 10 × 5.0

= 1.5×10-3 V

According to Fleming’s right hand rule, the direction of induced e.m.f is from west to east.

West and of the wire must be at higher electric potential.

Q:8.Current in a circuit falls from 5.0 A to 0.0 A in 01 s. if an average e.m.f of 200 V is induced, give an estimate of the self inductance of the circuit ?

Sol.
here , $\displaystyle \frac{dI}{dt} = \frac{I_2 – I_1}{t}$

$\displaystyle = \frac{0 – 5}{0.1} = 50 A/s$ ; e=200 V, L = ?

As $\displaystyle e = L |\frac{dI}{dt}|$

$\displaystyle L = \frac{e}{|\frac{dI}{dt}|}$

$\displaystyle L = \frac{200}{50}$

L = 4 H

Q: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 in flux linkage with the other coil ?

Sol.
Here, M =1.5 H, dI/dt =(20-0)/0.5 = 40 As-1

Form $\displaystyle e = M\frac{dI}{dt}$

$\displaystyle \frac{d\phi}{dt} = M\frac{dI}{dt}$

$\displaystyle d\phi = M dI$

$\displaystyle = 1.5 (20 – 0) = 30 wb$

Q:10. A jet plane is travelling west at the speed of 1800 km/h What is the voltage between the ends of the wing 25 m long, if the earth’s magnetic field at the location has a magnitude of 5.0 ×10-4 T and the dip angle is 30° .

Sol: v = 1800 km/h = 1800 x(5/18) = 500 m/s ; B = 5.0 ×10-4 T ; δ = 30°

$\displaystyle B_V = B sin\delta$

$\displaystyle e = B_V l v$

$\displaystyle e = (B sin\delta) l v$

$\displaystyle e = (5 \times 10^{-4} sin30) 25 \times 500$

e = 3.1 volt

Next Page →