A conducting circular loop of area $$1.0m^{2}$$ is placed perpendicular to a magnetic field which varies as $$B = \sin(100 t)$$ Tesla. If the resistance of the loop is $$100 \Omega$$, then the average thermal energy dissipated in the loop in one period is _______J.

A conducting circular loop of area $$A = 1.0$$ m$$^2$$ is in a magnetic field $$B = \sin(100t)$$ T with resistance $$R = 100 \Omega$$, and we seek the average thermal energy dissipated in one period.

By Faraday’s law, the induced EMF is given by $$ \varepsilon = -\frac{d\Phi}{dt} = -A\frac{dB}{dt} = -1.0 \times \frac{d}{dt}[\sin(100t)] = -100\cos(100t) \text{ V}. $$ The peak EMF is $$\varepsilon_0 = 100$$ V and the angular frequency is $$\omega = 100$$ rad/s.

The instantaneous power dissipated in the resistance can be written as $$ P = \frac{\varepsilon^2}{R} = \frac{100^2 \cos^2(100t)}{100} = 100\cos^2(100t) \text{ W}. $$

Since the time-average of $$\cos^2(\omega t)$$ over a complete period is $$\frac{1}{2}$$, the average power becomes $$ \overline{P} = \frac{\varepsilon_0^2}{2R} = \frac{(100)^2}{2 \times 100} = \frac{10000}{200} = 50 \text{ W}. $$

The period of oscillation is $$ T = \frac{2\pi}{\omega} = \frac{2\pi}{100} \text{ s}. $$ Multiplying the average power by the period gives the energy dissipated in one period as $$ E = \overline{P} \times T = 50 \times \frac{2\pi}{100} = \frac{100\pi}{100} = \pi \text{ J}. $$

The correct answer is Option (3): $$\pi$$ J