Suppose the loop in Exercise 6.4 is stationary but the current feeding

Solution:

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

Hence, area of the rectangular wire loop,

A = length × width

$=8 \times 2=16 \mathrm{~cm}^{2}$

$=16 \times 10^{-4} \mathrm{~m}^{2}$

Initial value of the magnetic field, $B^{\prime}=0.3 \mathrm{~T}$

Rate of decrease of the magnetic field, $\frac{d B}{d t}=0.02 \mathrm{~T} / \mathrm{s}$

Emf developed in the loop is given as:

$e=\frac{d \phi}{d t}$

Where,

$d \phi=$ Change in flux through the loop area

= AB

$\therefore e=\frac{d(A B)}{d t}=\frac{A d B}{d t}$

$=16 \times 10^{-4} \times 0.02=0.32 \times 10^{-4} \mathrm{~V}$

Resistance of the loop, R = 1.6 Ω

The current induced in the loop is given as:

$i=\frac{e}{R}$

$=\frac{0.32 \times 10^{-4}}{1.6}=2 \times 10^{-5} \mathrm{~A}$

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

$P=i^{2} R$

$=\left(2 \times 10^{-5}\right)^{2} \times 1.6$

$=6.4 \times 10^{-10} \mathrm{~W}$

The source of this heat loss is an external agent, which is responsible for changing the magnetic field with time.