A circular coil of resistance ' $R$ ', area ' $A$ ', number of turns ' N ' is rotated about its vertical…

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
  1. $\frac{\mathrm{N}^2 \mathrm{~A}^2 \mathrm{~B}^2 \omega^2}{2 \mathrm{R}}$
  2. $\frac{B N A \omega}{R}$
  3. $\frac{B N A \omega}{2 R}$
  4. $\frac{\mathrm{N}^2 \mathrm{~A}^2 \mathrm{~B}^2 \omega^2}{\mathrm{R}}$

Solution

For a circular coil, $\begin{aligned} & e_0=N A B \omega \\ & i_0=\frac{N A B \omega}{R} \end{aligned}$ $\begin{aligned} \therefore \quad & \text { Average power dissipated per cycle }=\frac{1}{2} \mathrm{e}_0 \mathrm{i}_0 \\ & =\frac{1}{2}(\mathrm{NAB} \omega) \frac{(\mathrm{NAB} \omega)}{R} \\ & =\frac{\mathrm{N}^2 \mathrm{~A}^2 \mathrm{~B}^2 \omega^2}{2 R} \end{aligned}$

Asked in: MHT CET 2024 (16 May Shift 2)

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