JEE MAIN - Physics (2016 - 10th April Morning Slot - No. 12)
A conducting metal circular-wire-loop of radius r is placed perpendicular to a
magnetic field which varies with time as
B = B0e$${^{{{ - t} \over r}}}$$ , where B0 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 (t $$ \to $$ $$\infty $$) is :
B = B0e$${^{{{ - t} \over r}}}$$ , where B0 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 (t $$ \to $$ $$\infty $$) is :
$${{{\pi ^2}{r^4}B_0^4} \over {2\tau R}}$$
$${{{\pi ^2}{r^4}B_0^2} \over {2\tau R}}$$
$${{{\pi ^2}{r^4}B_0^2R} \over \tau }$$
$${{{\pi ^2}{r^4}B_0^2} \over {\tau R}}$$
Explanation
Given,
B = B0e$$^{ - {t \over \tau }}$$
Area of the circular loop, A = $$\pi $$ r2
$$ \therefore $$ Flux $$\phi $$ = BA = $$\pi $$ r2 B0 e$$^{ - {t \over \tau }}$$
Induced emf in the loop,
$$\varepsilon $$ = $$-$$ $${{d\phi } \over {dt}}$$ = $$\pi $$ r2B0$${1 \over \tau }$$e$$^{ - {t \over \tau }}$$
Heat generated
= $$\int\limits_0^ \propto {{i^2}R\,dt} $$
= $$\int\limits_0^ \propto {{{{\varepsilon ^2}} \over R}} \,dt$$
= $${1 \over R}{{{\pi ^2}{r^4}B_0^2} \over {{\tau ^2}}}\int\limits_0^ \propto {{e^{ - {{2t} \over \tau }}}} \,dt$$
= $${{{\pi ^2}{r^4}B_0^2} \over {{\tau ^2}R}} \times {1 \over {\left( { - {2 \over \tau }} \right)}}\left[ {{e^{ - {{2t} \over \tau }}}} \right]_0^ \propto $$
= $${{ - {\pi ^2}{r^4}B_0^2} \over {2{\tau ^2}R}} \times \tau \left( {0 - 1} \right)$$
= $${{{\pi ^2}{r^4}B_0^2} \over {2\tau R}}$$
B = B0e$$^{ - {t \over \tau }}$$
Area of the circular loop, A = $$\pi $$ r2
$$ \therefore $$ Flux $$\phi $$ = BA = $$\pi $$ r2 B0 e$$^{ - {t \over \tau }}$$
Induced emf in the loop,
$$\varepsilon $$ = $$-$$ $${{d\phi } \over {dt}}$$ = $$\pi $$ r2B0$${1 \over \tau }$$e$$^{ - {t \over \tau }}$$
Heat generated
= $$\int\limits_0^ \propto {{i^2}R\,dt} $$
= $$\int\limits_0^ \propto {{{{\varepsilon ^2}} \over R}} \,dt$$
= $${1 \over R}{{{\pi ^2}{r^4}B_0^2} \over {{\tau ^2}}}\int\limits_0^ \propto {{e^{ - {{2t} \over \tau }}}} \,dt$$
= $${{{\pi ^2}{r^4}B_0^2} \over {{\tau ^2}R}} \times {1 \over {\left( { - {2 \over \tau }} \right)}}\left[ {{e^{ - {{2t} \over \tau }}}} \right]_0^ \propto $$
= $${{ - {\pi ^2}{r^4}B_0^2} \over {2{\tau ^2}R}} \times \tau \left( {0 - 1} \right)$$
= $${{{\pi ^2}{r^4}B_0^2} \over {2\tau R}}$$
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