A conducting circular loop of resistance 20 O | vedclass

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(C) The magnetic flux $\phi$ through the loop is given by $\phi = B \cdot A = 20 	imes 10^{-2} 	imes 2 \sin(50 \pi t) = 0.4 \sin(50 \pi t) \,Wb$. Th

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(C) The magnetic flux $\phi$ through the loop is given by $\phi = B \cdot A = 20 	imes 10^{-2} 	imes 2 \sin(50 \pi t) = 0.4 \sin(50 \pi t) \,Wb$. The induced current is $I = \frac{e}{R} = -\frac{1}{R} \frac{d\phi}{dt}$. The charge flowing through the loop is $\Delta q = \int_{0}^{t} I \,dt = -\frac{1}{R} \int_{\phi_1}^{\phi_2} d\phi = -\frac{\phi_2 - \phi_1}{R}$. At $t = 0$, $\phi_1 = 0.4 \sin(0) = 0 \,Wb$. At $t = 20 \,ms = 0.02 \,s$, $\phi_2 = 0.4 \sin(50 \pi 	imes 0.02) = 0.4 \sin(\pi) = 0 \,Wb$. Thus, the net charge $\Delta q = -\frac{0 - 0}{20} = 0 \,C$.

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