JEE Mains · Physics · STD 12 -6. Electromagnetic induction
A conducting metal circular-wire-loop of radius \(r\) is placed perpendicular to a magnetic field which varies with time as \(B = {B_0}{e^{\frac{{ - t}}{\tau }}}\) , 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 \(\left( {t \to \infty } \right)\) 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 \(\frac{{{\pi ^2}{r^4}B_0^2R}}{\tau }\)
- D \(\frac{{{\pi ^2}{r^4}B_0^2}}{{\tau R}}\)
Answer & Solution
Correct Answer
(B) \(\frac{{{\pi ^2}{r^4}B_0^2}}{{2\tau R}}\)
Step-by-step Solution
Detailed explanation
Electric flux is given by \(\phi=\mathrm{B} . \mathrm{A}\) \(\phi=\mathrm{B}_{0} \pi \mathrm{r}^{2} \mathrm{e}^{-1 / \tau}\) \(\left(\because B=B_{0} e^{-t / \tau}\right)\) Induced \(E.m.f.\)…
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