Shown in the figure is a circular loop of radius r and resistance R . A variable magnetic…
Physics · JEE Advanced · NTA Exams — Electromagnetic Induction and Alternating Currents
Shown in the figure is a circular loop of radius r and resistance R. A variable magnetic field of induction B = B0e–t is established inside the coil. If the key (K) is closed at t = 0 the electrical power developed right after closing the switch is equal to
- \(\frac{\mathrm{B}_{0}^{2} \pi \mathrm{r}^{2}}{\mathrm{R}}\)
- \(\frac{\mathrm{B}_{0} \emptyset \mathrm{r}^{3}}{\mathrm{R}}\)
- \(\frac{\mathrm{B}_{0}^{2} \pi^{2} \mathrm{r}^{4} \mathrm{R}}{5}\)
- \(\frac{\mathrm{B}_{0}^{2} \pi^{2} \mathrm{r}^{4}}{\mathrm{R}}\)
Answer
(D) B _ 0 ^ 2 ^ 2 r ^ 4 R
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