NCERT Questions for Class 12 Physics Chapter 6 – Electromagnetic Induction

Important Questions with Solutions of Class 12 Physics Chapter 6 – Electromagnetic Induction

1) Why does a metallic element become extremely hot when surrounded by a coil of high-frequency (H.F) alternate current?

Ans – If a metallic element is surrounded by a coil of high-frequency (H.F) alternate current, it will turn hot due to the creation of eddy currents responsible for producing the joule’s heating effect.

2) A round copper disc with a radius of 10cm rotates at a 2πrad/s speed about an axis from its centre & perpendicular to the disc. A uniform magnetic field of 0.2T remains perpendicular to the disc.

a) Find the potential difference created between the axis & rim of the disc.

Ans – As you know, radius = 10cm, B = 0.2T & ω = 2πrad/s

ε = 1/2 * BWr²

⇒ ε = 1/2 * 0.2 * 2π * (0.1)²

⇒ ε = 0.00628 volts

Hence, the potential difference developed is 0.00628 volts.

b) Predict the Induced current when the disc resistance is 2Ω!

Ans – I = ε/R = 0.0628/2

⇒ I = 0.0314A.

If the resistance is 2Ω, then the induced current will be 0.0314A.

3) A jet plane is moving over west with 1800km/hr in speed. Calculate the potential difference created between the wing ends of 25m distance, when the Earth’s magnetic field has a magnitude of 5 x 10^-4T & the dip angle is 30°.

Ans – As the velocity of the jet plane, v = 1800km/hr = 500m/s

Wingspan of jet plane, l = 25m

Strength of earth’s magnetic field, B = 5 x 10^-4T

The angle of dip, δ = 30

Vertical component of earth’s magnetic field, B_v = Bsinδ

⇒ B_v = 5 x 10^-4 x sin30°

⇒ 2.5 x 10^-4 T

Hence, the potential difference between the wing ends, e = (B_v) x I x v

⇒ e = 2.5 x 10^-4 x 25 x 500 = 3.125V

The voltage difference created between wing ends will be 3.125V.

4) A 10 m long straight wire lying from East to West is resting 5 ms^-1 in speed perpendicular to the horizontal component of the earth’s magnetic field 0.3 x 10^-4 Wbm^-2

(a) Find the instantaneous value of the induced EMF in the wire.

Ans – We all know the length of the wire, I = 10m

Falling speed of the wire, v = 5m/s

Magnetic field strength, B = 0.3 x 10^-4 Wbm^-2

EMF induced in the wire, e = BIv

⇒ e = 0.3 x 10^-4 x 5 x 10

⇒ e = 1.5 x 10^-3 V

Hence, the instantaneous EMF induced will be 1.5 x 10^-3 V.

b) In which direction is the induced emf?

Ans – The direction of the induced EMF travels from west to east according to Fleming’s right-hand rule.

c) Which end of the wire has the maximum electrical potential?

Ans – The maximum potential lies at the eastern end of the wire.

5) A 1.0ms lengthy metallic rod is rotated with an angular frequency of 400 rads^-1 in an axis normal to the rod over one end. The opposite rod end has a round metallic ring—a constant & constant magnetic field of 0.5T parallel to the axis. Find the EMF created between the centre & the ring.

Ans – As we know, the length of the rod, l = 1m

Angular frequency, ω = 400 rad/s

Magnetic field strength, B = 0.5T

One end of the rod has zero linear velocity while the other has a linear velocity of I_(t).

Average linear velocity of the rod,

EMF created between the centre & the ring,

Thus, the EMF created between the centre & the ring is 100V.

6) Differentiate between resistances, reactance & impedance of an AC circuit.

Ans – It can be segregated as:

Resistance	Reactance	Impedance
Opposition provided by the resistor to the current flow.	Opposition is provided by the inductor/capacitor to the current flow.	The combination of resistor, inductor & capacitor provides opposition.
It’s free of source frequency.	It’s based on the source frequency.	It’s based on the source frequency.

7) Two coils placed adjacent have a mutual inductance of 1.5H. When the current through one coil varies from 0 to 20A in 0.5s, calculate the difference in flux linkage associated with the other coil.

Ans – As we know, mutual inductance of a pair of coils, μ = 1.5H

Initial current, I₁ = 0A

Final current, I₂ = 20A

Change in current, dI = I₂ – I₁ = 20 – 0 = 20A

Time for the change, dt = 0.5s

Induced EMF, e = dφ/dt …. (1)

Here dφ is the transformation in the flux linkages with the coil.

EMF is related with mutual inductance, e = µ * dl/dt …. (2)

Equating eq (1) & (2)

⇒ dφ = 1.5 x 20

⇒ dφ = 30Wb

Hence, the change in flux linkage is 30Wb.

8) A conducting rod moves with angular speed w having one end at the centre & another end at the circumference of a round metallic ring with radius R where an axis passes via the coil centre perpendicular to the coil plane A & constant magnetic field B parallel to the axis is visible all over. Calculate the EMF between the centre & the metallic ring defined by 1/2BwR²

Ans – When a circular loop joins the centre with point P having a resistor. Then, the Potential difference over the resistor = induced EMF

i.e, ε = B * Rate of change in the loop area

When resistor QP is shifted with angular velocity ω & moves by an angle θ in time t then,

Area cleared,

Hence, φ = BA cosθ = BA

So,

Hence, it’s been clarified that the EMF between the centre & the metallic ring can be ½ BωR².

9) A rectangular wire loop with 8cm & 2cm sides of small cut moves from a region of uniform magnetic field at 0.3T magnitude pointed towards the loop. Calculate the EMF created over the cut when the loop velocity is 1cm/s in a position normal to the

(a) Extended Region:

Ans – As we know, the length of the rectangular wire, l = 8cm = 0.08 m

Width of the rectangular wire, b = 2cm = 0.02 m

Area of the rectangular loop: A = Ib = 0.09 x 0.02

= 16 x 10^-4 m²

Magnetic field strength, B = 0.3T

Loop velocity, v = 1cm/s = 0.01m/s

EMF created in the loop can be, e = Blv

=> 0.3 x 0.08 x 0.01 = 2.4 x 10^-4 V

The time taken to travel along the width will be,

t = Distance Travelled/Velocity = b/v = 0.02/0.01 = 2s

Hence, the final induced voltage will be 2.4 x 10^-4 V, lasting 2s.

b) In the shorter loop side – Calculate the induced voltage duration in each case:

Ans – EMF created, e = Bbv

=> 0.3 x 0.02 x 0.01 = 0.6 x 10^-4 V

Time taken to travel,

t = Distance Traveled/Velocity = 1/v = 0.08/0.01 = 8s,

Hence, the final induced voltage remains 0.6 x 10^-4 V, lasting 8s.

10) A round-shaped coil with an 8.0cm radius & 20 turns is rotated over its vertical diameter with 50rads^-1 angular speed in a constant horizontal magnetic field with 3 x 10^-2 T magnitude. Calculate the maximum & average EMF induced in the coil. Find the maximum current coil value when the coil develops from a closed loop 100 in resistance. Estimate the average loss of power on the Joule heating phenomenon. What’s the source of this power?

Ans – As we know,

Maximum induced EMF = 0.603V

Average induced EMF = OV

Current in coil (Max) = 0.0603A

Average power loss = 0.018W (External rotor power)

Circular coil radius r = 8cm = 0.08m

Coil area A = πr² = π*(0.08)²m²

No. of coil turns N = 20

Angular speed ω = 50rad/s

Magnetic field resistance B = 3 x 10^-2T

Loop resistance R = 10Ω

Max-induced EMF will be,

ε = NωAB = 20 x 50 x π x (0.08)² x 3 x 10^-2 = 0.603V

Hence, the max EMF induced in the coil can be 0.603V.

The average EMF induced in the coil is zero during one full cycle.

Max current can be:

I = ε/R

⇒ I = 0.603/10 = 0.0603A

Average power loss during joule heating,

Since we understood that the induced current in the coil can create a torque opposite to the coil rotation, the rotor is an external agent. It offers torque against this torque to maintain the coil rotating uniformly. So, the external rotor delivers the dissipated power.