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A circular coil of area $0.01 \mathrm{~m}^2$ and 40 turns is rotated about its vertical diameter with an angular speed of 50 $\mathrm{rad} \mathrm{s}^{-1}$ in a uniform horizontal magnetic field $0.05 \mathrm{~T}$. If the average power loss due to joule heating is $25 \mathrm{~mW}$, then the closed loop resistance of the coil is
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Verified Answer
The correct answer is:
$20 \Omega$
We have
$\begin{aligned} & \mathrm{e}=\frac{\mathrm{d}}{\mathrm{dt}}=\mathrm{NBA} \frac{\mathrm{d}}{\mathrm{dt}}(\sin \omega \mathrm{t}) \\ & =\mathrm{NBA} \omega \sin \omega \mathrm{t}\end{aligned}$
$\begin{aligned} & \mathrm{e}_{\max }=\mathrm{NBA} \omega \\ & \mathrm{e}_{\mathrm{rms}}=\frac{\mathrm{NBA} \omega}{\sqrt{2}}\end{aligned}$
$\begin{aligned} & P_{\mathrm{avg}}=\frac{\mathrm{e}_{\mathrm{rms}}^2}{\mathrm{R}} \Rightarrow \mathrm{R}=\frac{\mathrm{e}_{\mathrm{rms}}^2}{\mathrm{P}_{\mathrm{avg}}}=\frac{\mathrm{N}^2 \mathrm{~B}^2 \mathrm{~A}^2 \omega^2}{2 \times \mathrm{P}_{\mathrm{avg}}} \\ & =\frac{40^2 \times 0.05^2 \times 0.01^2 \times 50^2}{2 \times 25 \times 10^{-3}} \\ & =20 \Omega\end{aligned}$
$\begin{aligned} & \mathrm{e}=\frac{\mathrm{d}}{\mathrm{dt}}=\mathrm{NBA} \frac{\mathrm{d}}{\mathrm{dt}}(\sin \omega \mathrm{t}) \\ & =\mathrm{NBA} \omega \sin \omega \mathrm{t}\end{aligned}$
$\begin{aligned} & \mathrm{e}_{\max }=\mathrm{NBA} \omega \\ & \mathrm{e}_{\mathrm{rms}}=\frac{\mathrm{NBA} \omega}{\sqrt{2}}\end{aligned}$
$\begin{aligned} & P_{\mathrm{avg}}=\frac{\mathrm{e}_{\mathrm{rms}}^2}{\mathrm{R}} \Rightarrow \mathrm{R}=\frac{\mathrm{e}_{\mathrm{rms}}^2}{\mathrm{P}_{\mathrm{avg}}}=\frac{\mathrm{N}^2 \mathrm{~B}^2 \mathrm{~A}^2 \omega^2}{2 \times \mathrm{P}_{\mathrm{avg}}} \\ & =\frac{40^2 \times 0.05^2 \times 0.01^2 \times 50^2}{2 \times 25 \times 10^{-3}} \\ & =20 \Omega\end{aligned}$
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