Question
A conducting circular loop of area $1.0{\text{ }m}^{2}$ is placed perpendicular to a magnetic field which varies as $B = sin(100t)$ Tesla. If the resistance of the loop is $100\Omega$, then the average thermal energy dissipated in the loop in one period is $\_\_\_\_$ J.
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Solution
Area of the loop $= 1{\text{ }m}^{2}$
$${B = sin(100t) }{\therefore\ \phi = BA = sin(100t) }{\therefore\ \frac{d\phi}{dt} = 100cos(100t) }{\therefore\ P = \frac{V^{2}}{R} = \frac{10^{4}\cos^{2}(100t)}{100} }$$$\therefore\ $ Thermal energy dissipated in 1 time period
$$\begin{matrix} & \ = \int_{0}^{T}\mspace{2mu}\mspace{2mu} Pdt = \int_{0}^{T}\mspace{2mu}\mspace{2mu} 100\cos^{2}(100t)dt \\ & \text{ }T = \frac{2\pi}{100} = \frac{\pi}{50}\sec \\ & \ \therefore\ Q = 100\int_{0}^{\pi/50}\mspace{2mu}\mspace{2mu}\cos^{2}(100t)dt \\ & \ = 100\int_{0}^{\pi/50}\mspace{2mu}\mspace{2mu}\frac{1 + cos200t}{2}dt \\ & \ = 100\left\lbrack \frac{\pi}{100} \right\rbrack = \pi \end{matrix}$$
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