Concept:
As the coil rotates in a magnetic field, it induces an alternating e.m.f. given by Faraday's law. The peak induced e.m.f. ($E_0$) is $E_0 = NAB\omega$.
The question asks for the average power dissipated (often termed simply as "power dissipated" in standard AC circuit contexts). The average power $P$ for a purely resistive circuit is $P = E_{rms} \times I_{rms} = \frac{E_0}{\sqrt{2 \times \frac{I_0}{\sqrt{2 = \frac{E_0^2}{2R}$.
Step 1: Calculate the peak induced e.m.f. ($E_0$).
Given data:
Number of turns, $N = 100$
Area, $A = 2 \text{ m}^2$
Magnetic field, $B = 2 \times 10^{-2} \text{ T}$
Angular speed, $\omega = 30 \text{ rad/s}$
Use the formula for peak e.m.f.:
$$E_0 = NAB\omega$$
$$E_0 = 100 \times 2 \times (2 \times 10^{-2}) \times 30$$
$$E_0 = 100 \times 10^{-2} \times 2 \times 2 \times 30$$
$$E_0 = 1 \times 4 \times 30 = 120 \text{ V}$$
Step 2: Calculate the average power dissipated.
Given the circuit resistance $R = 600 \text{ }\Omega$.
The peak current is $I_0 = \frac{E_0}{R} = \frac{120}{600} = 0.2 \text{ A}$.
The average power dissipated over a full cycle in an AC generator circuit is:
$$P = \frac{E_0 I_0}{2}$$
$$P = \frac{120 \times 0.2}{2}$$
$$P = \frac{24}{2} = 12 \text{ W}$$
(Note: Standard competitive exam phrasing sometimes uses "maximum power dissipated" to refer to the average power capacity of the generator, distinguished from instantaneous peak power $E_0^2/R = 24 \text{ W}$).