In a coil of resistance 8 , Omega,the magneti | vedclass

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(C) Given,$\phi = \frac{2}{3}(9 - t^2)$.The flux becomes zero when $\phi = 0$,so $\frac{2}{3}(9 - t^2) = 0$,which gives $t^2 = 9$,or $t = 3 \, 	ext{s

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$2$

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(C) Given,$\phi = \frac{2}{3}(9 - t^2)$.The flux becomes zero when $\phi = 0$,so $\frac{2}{3}(9 - t^2) = 0$,which gives $t^2 = 9$,or $t = 3 \, 	ext{s}$.The induced electromotive force $(EMF)$ is given by Faraday's law: $e = -\frac{d\phi}{dt}$.$e = -\frac{d}{dt} [\frac{2}{3}(9 - t^2)] = -\frac{2}{3}(0 - 2t) = \frac{4t}{3}$.The heat produced $H$ in the coil is given by $H = \int_{0}^{t} \frac{e^2}{R} dt$.Substituting the values $e = \frac{4t}{3}$ and $R = 8 \, \Omega$:$H = \int_{0}^{3} \frac{(\frac{4t}{3})^2}{8} dt = \int_{0}^{3} \frac{16t^2}{9 	imes 8} dt = \int_{0}^{3} \frac{2t^2}{9} dt$.$H = \frac{2}{9} [\frac{t^3}{3}]_{0}^{3} = \frac{2}{9} 	imes \frac{27}{3} = \frac{2}{9} 	imes 9 = 2 \, 	ext{J}$.

In a coil of resistance $8 \, \Omega$,the magnetic flux due to an external magnetic field varies with time as $\phi = \frac{2}{3}(9 - t^2)$. The value of total heat produced in the coil,till the flux becomes zero,will be $.... \, J$.

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