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  • Shown in the figure is a circular loop of radius r and resistance R. A variable magnetic field of induction    <img alt="\mathrm{B=B_0 e^{-t}}" src="https://entrancecorner.oncodecogs

Shown in the figure is a circular loop of radius r and resistance R. A variable magnetic field of induction    \mathrm{B=B_0 e^{-t}} is established inside the coil. If the key (K) is closed, the electrical power developed right after closing the switch is equal to

Option: 1

\mathrm{\frac{B_0^2 \pi r^2}{R}}


Option: 2

\mathrm{\frac{B_{0 }\pi r^3}{R}}


Option: 3

\mathrm{\frac{B_0^2 \pi^2 r^4 R}{R}}


Option: 4

\mathrm{\frac{B_0^2 \pi^2 r^4}{R}}


Answers (1)

best_answer

\mathrm{\text { Then induced emf }=|\varepsilon|=\frac{\mathrm{d} \phi}{\mathrm{dt}}}

\mathrm{\begin{aligned} & =\frac{d}{d t}(B \cdot A) \quad=A \frac{d B}{d t} \\ & =\left(\pi r^2\right) B_0 \frac{d}{d t}\left(\bar{e}^t\right)=-\pi r^2 B_0 \bar{e}^t \\ & \Rightarrow \varepsilon_0=B_0 \pi r^2 \end{aligned}}

The electrical power developed in the resister just at the instant of closing the key \mathrm{=P=\frac{\varepsilon_0^2}{R}=\frac{B_0^2 \pi^2 r^4}{R}}

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Gunjita

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