**Question:**

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?

**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.

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