Sides of the rectangular loop are 8 cm and 2 cm.

Hence, area of the rectangular wire loop,

A = length x width = 8 x 2 = 16 cm² = 16 x 10⁻⁴ m²

Initial value of the magnetic field, B’ = 0.3 T

Rate of decrease of the magnetic field, dB/dt = 0.02 T/s

Emf developed in the loop is given as: e = dφ/dt

Where, dφ= Change in flux through the loop area = AB

Therefore, e = d(AB)/dt = AdB/dt

= 16 x 10⁻⁴ x 0.02 = 0.32 x 10⁻⁴ V

Resistance of the loop, R = 1.6 Ω

The current induced in the loop is given as: i = e/R

= (0.32 x 10⁻⁴)/1.6 = 2 x 10⁻⁵ A

Power dissipated in the loop in the form of heat is given as:

P = i²R

= (2 x 10⁻⁵)² x 1.6 = 6.4 x 10⁻¹⁰ W
The source of this heat loss is an external agent, which is responsible for changing the magnetic field with time.

Mutual inductance of a pair of coils, μ = 1.5 H

Initial current, I₁= 0 A Final current I₂ = 20 A

Change in current, dl = I₂ — I₁= 20 — 0 = 20 A

Time taken for the change, t = 0.5 s

Induced emf e = dφ/dt ——————Eq-1

Where, dφ is the change in the flux linkage with the coil.

Emf is related with mutual inductance as:

e = μ dI /dt —————————-Eq-2

Equating equations (1) and (2), we get

dφ/dt = μ dI /dt

dφ = 1.5x(20)

= 30 Wb

Hence, the change in the flux linkage is 30 Wb.

Length of the wire, 1 = 10 m

Falling speed of the wire, v = 5.0 m/s

Magnetic field strength, B = 0.3 x 10^-4 Wb m^–2

Ans (a).

Emf induced in the wire, e = Blv = 0.3 x 10^-4 x 5 x 10 = 1.5 x 10^-3 V

Ans (b).

Using Fleming’s right hand rule, it can be inferred that the direction of the induced emf is from West to East.

Ans (c).

The eastern end of the wire is at a higher potential.