A conducting circular loop of area $1.0 \mathrm{~m}^2$ is placed perpendicular to a magnetic field w

Question

A conducting circular loop of area $1.0 \mathrm{~m}^2$ 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.

Professor PC Thomas Classes · Answer

Magnetic flux $\phi = BA = (1.0)\sin(100t) = \sin(100t)$. The induced emf is $e = -\frac{d\phi}{dt} = -100\cos(100t)$. The instantaneous power dissipated is $P = \frac{e^2}{R} = \frac{100^2 \cos^2(100t)}{100} = 100\cos^2(100t)$. The average power over one cycle is $P_{avg} = 100 	imes \frac{1}{2} = 50 	ext{ W}$. The time period is $T = \frac{2\pi}{\omega} = \frac{2\pi}{100} = \frac{\pi}{50} 	ext{ s}$. Therefore, average thermal energy dissipated in one period is $E = P_{avg} 	imes T = 50 	imes \frac{\pi}{50} = \pi 	ext{ J}$.