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

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. $\frac{\mathrm{BNA} \omega}{\mathrm{R}}$
  3. $\frac{N^2 A B}{2 R \omega^2}$
  4. $\frac{\mathrm{BA} \omega}{2 \mathrm{NR}}$

Solution

Average power dissipated in an AC circuit is, $P_{a v}=\frac{e_0 i_0}{2}$
But, $\mathrm{i}_0=\frac{\mathrm{e}_0}{\mathrm{R}} \Rightarrow \mathrm{P}_{\mathrm{av}}=\frac{\mathrm{e}_0^2}{2 \mathrm{R}}$ For the given circular coil, $\begin{aligned} & \mathrm{e}_0=\mathrm{NAB} \omega \\ \therefore \quad & \mathrm{P}_{\mathrm{av}}=\frac{\mathrm{N}^2 \mathrm{~B}^2 \mathrm{~A}^2 \omega^2}{2 \mathrm{R}} \end{aligned}$

Asked in: MHT CET 2024 (04 May Shift 1)

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