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A conducting metal circular-wire-loop of radius r is placed perpendicular to a magnetic field which varies with time as
$B = B_{0}e^{-t/\tau } ,$ 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 $\left ( t\rightarrow \infty \right )$ is :

Option 1)
$\frac{\pi ^{2}r^{4}B_{0}^{4}}{2\tau R}$

Option 2)
$\frac{\pi ^{2}r^{4}B_{0}^{2}}{2\tau R}$

Option 3)
$\frac{\pi ^{2}r^{4}B_{0}^{2} R}{\tau }$

Option 4)
$\frac{\pi ^{2}r^{4}B_{0}^{2} }{\tau R }$

Answers (1)

best_answer

As we have learned

Induced Current -

$I= \frac{\varepsilon }{R}=\frac{-N}{R}\frac{d\phi }{dt}$

- wherein

$R\rightarrow$ Resistance

$\frac{d\phi }{dt}\rightarrow$ Rate of change of flux

$B = B_0 e^{ -t / \tau }$

Heat generated = $\int_{0}^{\infty }i^2 Rdt$

$\int_{0}^{\infty } \frac{\varepsilon _{ind}^2}{Rt}Rdt = \frac{1}{R } \int_{0}^{\infty } {\varepsilon _{ind}^2}dt$

$\frac{1}{R } \frac{\pi ^2 r^ 4 B_{0}^{2}}{\tau ^2 }\cdot \int_{0}^{\infty }e^{-2t/\tau }dt$

$= \frac{\pi ^2 r^4B_{0}^{2}}{R\tau ^2}\frac{e^{-2t/\tau }}{-2/\tau }|_{0}^{\infty }$

$H = \frac{\pi ^2 B_{0}^{2}r^4}{2 R \tau }$

Option 1)

$\frac{\pi ^{2}r^{4}B_{0}^{4}}{2\tau R}$

Option 2)

$\frac{\pi ^{2}r^{4}B_{0}^{2}}{2\tau R}$

Option 3)

$\frac{\pi ^{2}r^{4}B_{0}^{2} R}{\tau }$

Option 4)

$\frac{\pi ^{2}r^{4}B_{0}^{2} }{\tau R }$

Posted by

Avinash

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