The coil of an a.c. generator has 100 turns, each of cross-sectional area $2 \mathrm{~m}^2$. It is rotating…
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
$6\ W$
$9\ W$
$12\ W$
$24\ W$
Solution
$\begin{aligned}
& \mathrm{N}=100, \mathrm{~A}=2 \mathrm{~m}^2, \omega=30 \mathrm{rad} / \mathrm{s} \\
& \mathrm{B}=2 \times 10^{-2} \mathrm{~T}, \mathrm{R}=600 \Omega
\end{aligned}$
Maximum power dissipated in the circuit
$\begin{aligned}
P_{\max }=E_{r m s} \times I_{r m s} & =\frac{E_0}{\sqrt{2}} \times \frac{I_0}{\sqrt{2}} \\
& =\frac{E_0 I_0}{2}...(i)
\end{aligned}$
But $\mathrm{I}_0=\frac{\mathrm{E}_0}{\mathrm{R}}...(ii)$
Putting (2) into (1) we get,
$\mathrm{I}_0=\frac{\mathrm{E}_0{ }^2}{2 \mathrm{R}}$
$\begin{aligned}
\text { But } \mathrm{E}_0 & =\mathrm{NAB} \omega \\
\mathrm{E}_0 & =100 \times 2 \times 2 \times 10^{-2} \times 30 \\
& =120 \mathrm{~V} \\
\therefore \quad \mathrm{P}_{\max }= & \frac{120 \times 120}{2 \times 600}=12 \mathrm{~W}
\end{aligned}$