6.11 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?

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Answered by

Payal Gupta | Contributor-Level 10

4 months ago

6.11 The area of the rectangular coil, A = 8 $\times 2 =$ 16 ${c m}^{2}$ = 16 $\times 10^{- 4}$ $m^{2}$

Initial value of the magnetic field, $B_{1}$ = 0.3 T

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

From the relation of induced emf e = $\frac{d \emptyset}{d t}$ , where $d \emptyset$ is the change in the flux linkage with the coil = A $\times B$

Hence, e = $\frac{d (A B)}{d t}$ = A $\frac{d B}{d t}$ = 16 $\times 10^{- 4} \times 0.02$ = 3.2 $\times 10^{- 5}$ V

Resistance in the loop, R = 1.6 Ω

Hence I = $\frac{e}{R}$ = $\frac{3.2 \times 10^{- 5}}{1.6}$ = 2 $\times 10^{- 5}$ A

Power dissipated in the form of heat is given by

P = $I^{2}$ R = ( ${2 \times 10^{- 5})}^{2} \times 1.6$ = 6.4 $\times 10^{- 10}$ W

The source of heat loss is an external agent, which is resp

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