The coil of an a.c. generator has 100 turns, each of cross-sectional area \(2 \mathrm{~m}^2\). It is rotating at constant angular speed \(30 \mathrm{rad} / \mathrm{s}\), in a uniform magnetic field of \(2 \times 10^{-2} \mathrm{~T}\). If the total resistance of the circuit is \(600 \Omega\) then maximum power dissipated in the circuit is

\(\begin{aligned}
& \mathrm{N}=100, \mathrm{~A}=2 \mathrm{~m}^2, \omega=30 \mathrm{rad} / \mathrm{s} \\
& \mathrm{B}=2 \times 10^{-2} \mathrm{~T}, \mathrm{R}=600 \Omega
\end{aligned}\)
Maximum power dissipated in the circuit
\(P_{\max }=E_{r m s} \times I_{r m s} =\frac{E_0}{\sqrt{2}} \times \frac{I_0}{\sqrt{2}}\)
\(=\frac{E_0 I_0}{2}...(i)\)
But \(\mathrm{I}_0=\frac{\mathrm{E}_0}{\mathrm{R}}...(ii)\)
Putting (2) into (1) we get,
\(\mathrm{I}_0=\frac{\mathrm{E}_0{ }^2}{2 \mathrm{R}}\)
\(\text {But } \mathrm{E}_0 =\mathrm{NAB} \omega\)
\(\mathrm{E}_0 =100 \times 2 \times 2 \times 10^{-2} \times 30\)
\(=120 \mathrm{~V}\)
\(\therefore \mathrm{P}_{\max }= \frac{120 \times 120}{2 \times 600}=12 \mathrm{~W}\)