MHT CET · Physics · 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
- A \(\frac{\mathrm{N}^2 \mathrm{~A}^2 \mathrm{~B}^2 \omega^2}{2 \mathrm{R}}\).
- B \(\frac{\mathrm{BNA} \omega}{\mathrm{R}}\)
- C \(\frac{N^2 A B}{2 R \omega^2}\)
- D \(\frac{\mathrm{BA} \omega}{2 \mathrm{NR}}\)
Answer & Solution
Correct Answer
(A) \(\frac{\mathrm{N}^2 \mathrm{~A}^2 \mathrm{~B}^2 \omega^2}{2 \mathrm{R}}\).
Step-by-step Solution
Detailed explanation
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,
\(\begin{aligned}
& \mathrm{e}_0=\mathrm{NAB} \omega \\
\therefore \quad & \mathrm{P}_{\mathrm{av}}=\frac{\mathrm{N}^2 \mathrm{~B}^2 \mathrm{~A}^2 \omega^2}{2 \mathrm{R}}
\end{aligned}\)
\(P_{a v}=\frac{e_0 i_0}{2}\)
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,
\(\begin{aligned}
& \mathrm{e}_0=\mathrm{NAB} \omega \\
\therefore \quad & \mathrm{P}_{\mathrm{av}}=\frac{\mathrm{N}^2 \mathrm{~B}^2 \mathrm{~A}^2 \omega^2}{2 \mathrm{R}}
\end{aligned}\)
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