A conducting metal circular-wire-loop of radius r is placed perpendicular to a magnetic field which varies with time as B = B0e (-t)/(τ), where B0 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: (a) π²r⁴B₀⁴2τ R (b) π²r⁴B₀²2τ R (c) π²r⁴B₀²Rτ (d) π²r⁴B₀²τ R

Answer: (b). e = -d/dt(π r² B₀ e-t/τ) = (π r² B₀)/(τ)e-t/τ. H = ∫₀∞ e²/Rdt = (π² r⁴ B₀²)/(τ² R)∫₀∞ e-2t/τdt. ∫₀∞ e-2t/τdt = τ/2. H = (π² r⁴ B₀²)/(2τ R). So the answer is (b).

A conducting metal circular-wire-loop of radius r is placed… | JEE Main 2016 (10 Apr, Shift 1)

Q77JEE Main 2016 · 10 APR · SHIFT 1Difficult

Magnetic Flux & Faraday's Law

A conducting metal circular-wire-loop of radius r is placed perpendicular to a magnetic field which varies with time as

B = B0e $\frac{-t}{\tau}$ , 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 (t $\to \infty$ ) is:

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Answer: (b)

$e = -\dfrac{d}{dt}(\pi r^2 B_0 e^{-t/\tau}) = \dfrac{\pi r^2 B_0}{\tau}e^{-t/\tau}$ .
$H = \int_0^\infty \dfrac{e^2}{R}dt = \dfrac{\pi^2 r^4 B_0^2}{\tau^2 R}\int_0^\infty e^{-2t/\tau}dt$ .
$\int_0^\infty e^{-2t/\tau}dt = \dfrac{\tau}{2}$ .
$H = \dfrac{\pi^2 r^4 B_0^2}{2\tau R}$ . So the answer is (b).

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Magnetic Flux & Faraday's Law question from JEE Main 2016 (10 Apr, Shift 1)

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