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Q.

Magnetic flux through a stationary loop with a resistance R varies during the time of interval $τ$ as $ϕ = a t (τ - t)$ where $a$ is a constant. The amount of heat generated in the loop during the time interval $τ$ is

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a

$\frac{a^{2} τ^{3}}{6 R}$

b

$\frac{a^{2} τ^{3}}{4 R}$

c

$\frac{a^{2} τ^{3}}{3 R}$

d

$\frac{a^{2} τ^{3}}{2 R}$

answer is C.

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Detailed Solution

$e = \frac{d ϕ}{d t} = a τ - 2 a t$

$\frac{d H}{d t} = \frac{e^{2}}{R} = \frac{{(a τ - 2 a t)}^{2}}{R}$

$H = \int_{0}^{τ} \frac{{(a τ - 2 a t)}^{2}}{R} d t = {[\frac{{(a τ - 2 a t)}^{3}}{R (3) (- 2 a)}]}_{0}^{τ} = \frac{{(a τ - 2 a τ)}^{3} - {(a τ)}^{3}}{(- 6 a) R} = \frac{- a^{3} τ^{3} - a^{3} τ^{3}}{(- 1) (6 a) R} = \frac{a^{3} τ^{3}}{3 R}$

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