A conducting metal circular-wire-loop of radius r is placed perpendicular to a magnetic field which varies with time as B = B0 e^ - t, where B0 and are constants, at time t = 0. If the resistance of the loop is R then the heat generated in the loop after a long time ( t →) is

(A) \frac{{{\pi ^2}{r^4}B_0^4}}{{2\tau R}}

(C) \frac{{{\pi ^2}{r^4}B_0^2R}}{\tau }

(D) \frac{{{\pi ^2}{r^4}B_0^2}}{{\tau R}}

JEE Mains · Physics · STD 12 -Electromagnetic induction

A conducting metal circular-wire-loop of radius \(r\) is placed perpendicular to a magnetic field which varies with time as \(B = {B_0}{e^{\frac{{ - t}}{\tau }}}\) , where \(B_0\) and \(\tau \) are constants, at time \(t = 0\). If the resistance of the loop is \(R\) then the heat generated in the loop after a long time \(\left( {t \to \infty } \right)\) is

Pick an option to check your answer.

Verified Solution

Correct Answer

(B) \(\frac{{{\pi ^2}{r^4}B_0^2}}{{2\tau R}}\)

Detailed explanation

Electric flux is given by

\(\phi=\mathrm{B} . \mathrm{A}\)

\(\phi=\mathrm{B}_{0} \pi \mathrm{r}^{2} \mathrm{e}^{-1 / \tau}\)

\(\left(\because B=B_{0} e^{-t / \tau}\right)\)

Induced \(E.m.f.\) \(\varepsilon=\frac{\mathrm{d} \phi}{\mathrm{dt}}=\frac{\mathrm{B}_{0} \pi \mathrm{r}^{2}}{\tau^{2}} \mathrm{e}^{-t / \tau}\)

Heat \(=\int_{0}^{\infty} \frac{\varepsilon^{2}}{\mathrm{R}}=\frac{\pi^{2} \mathrm{r}^{4} \mathrm{B}_{0}^{2}}{2 \tau \mathrm{R}}\)

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