Question

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 \(B_0\) 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 \((t \to \infty)\) is :

• A. \(\frac{\pi^2 r^4 B_0^4}{2\tau R}\)
• B. \(\frac{\pi^2 r^4 B_0^2}{2\tau R}\)
• C. \(\frac{\pi^2 r^4 B_0^2 R}{\tau}\)
• D. \(\frac{\pi^2 r^4 B_0^2}{\tau R}\)

Hint:

Find emf from Faraday's law, then integrate \(\varepsilon^2/R\) from 0 to infinity.

Answer 1

The Correct Option is B

Flux \(\Phi = B\pi r^2 = \pi r^2 B_0 e^{-t/\tau}\).
EMF \(\varepsilon = -\frac{d\Phi}{dt} = \frac{\pi r^2 B_0}{\tau}e^{-t/\tau}\).
Heat \(H = \int_0^\infty \frac{\varepsilon^2}{R}dt = \frac{\pi^2 r^4 B_0^2}{\tau^2 R}\int_0^\infty e^{-2t/\tau}dt = \frac{\pi^2 r^4 B_0^2}{\tau^2 R}\cdot\frac{\tau}{2}\).
\(H = \frac{\pi^2 r^4 B_0^2}{2\tau R}\).
Answer: (2) \(\frac{\pi^2 r^4 B_0^2}{2\tau R}\)