JEE Main 2016 · 10 Apr · Shift 1 · Physics Q17

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PhysicsSingle correct+4 / -1Question 17 of 90

A conducting metal circular-wire-loop of radius r is placed perpendicular to a magnetic field which varies with time as $B=B_0e^{-t/\tau}$ , where $B_0$ and τ 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 :

\dfrac{\pi^2r^4B_0^4}{2\tau R}

\dfrac{\pi^2r^4B_0^2}{2\tau R}

\dfrac{\pi^2r^4B_0^2R}\tau

\dfrac{\pi^2r^4B_0^2}{\tau R}

Answer and worked solution

Official answer: B

$\varepsilon=\pi r^2\frac{B_0}\tau e^{-t/\tau}$ , so
$H=\int_0^\infty\frac{\varepsilon^2}Rdt=\frac{\pi^2r^4B_0^2}{\tau^2R}\cdot\frac\tau2=\frac{\pi^2r^4B_0^2}{2\tau R}.$

Topics:Faraday's and Lenz's laws

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JEE Main 2016 · 10 Apr · Shift 1, Physics question 17