Question 233: Electromagnetic Induction

Question

A circular coil of resistance ' $R$ ', area ' $A$ ', number of turns ' N ' is rotated about its vertical diameter with angular speed ' $\omega$ ' in a uniform magnetic field of magnitude ' $B$ '. The average power dissipated in a complete cycle is

Accepted Answer

$\begin{aligned} & 	ext { For a circular coil, } \ & \mathrm{e}_0=\mathrm{NAB} \omega \ & \mathrm{i}_0=\frac{\mathrm{NAB} \omega}{\mathrm{R}} \ & 	herefore \quad 	ext { Average power dissipated per cycle }=\frac{1}{2} \mathrm{e}_0 \mathrm{i}_0 \ & =\frac{1}{2}(\mathrm{NAB} \omega) \frac{(\mathrm{NAB} \omega)}{\mathrm{R}} \ & \quad=\frac{\mathrm{N}^2 \mathrm{~A}^2 \mathrm{~B}^2 \omega^2}{2 \mathrm{R}}\end{aligned}$

Answer

$\frac{\mathrm{N}^2 \mathrm{~A}^2 \mathrm{~B}^2 \omega^2}{2 \mathrm{R}}$

Answer

$\frac{\mathrm{BNA} \omega}{\mathrm{R}}$

Answer

$\frac{\mathrm{BNA} \omega}{2 \mathrm{R}}$

Answer

$\frac{N^2 A^2 B^2 \omega^2}{R}$

Electromagnetic Induction

A circular coil of resistance ' R ', area ' A ', number of turns ' N ' is rotated about its vertical diameter with angular speed ' \omega ' in a uniform magnetic field of magnitude ' B '. The average power dissipated in a complete cycle is

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