The current in an inductor is given by I = (3 | vedclass

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(B) Given,the current in the inductor is $I = (3t + 8) \ A$.The induced emf $\varepsilon$ in an inductor is given by the formula $\varepsilon = -L \fr

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$4$

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(B) Given,the current in the inductor is $I = (3t + 8) \ A$.The induced emf $\varepsilon$ in an inductor is given by the formula $\varepsilon = -L \frac{dI}{dt}$.The magnitude of the induced emf is $|\varepsilon| = L \left| \frac{dI}{dt} ight|$.Given $|\varepsilon| = 12 \ mV = 12 	imes 10^{-3} \ V$.First,calculate the rate of change of current: $\frac{dI}{dt} = \frac{d}{dt}(3t + 8) = 3 \ A/s$.Now,substitute the values into the formula: $12 \ mV = L 	imes 3 \ A/s$.$L = \frac{12 \ mV}{3 \ A/s} = 4 \ mH$.Therefore,the self-inductance of the inductor is $4 \ mH$.

The current in an inductor is given by $I = (3t + 8)$ $A$,where $t$ is in seconds. The magnitude of the induced emf produced in the inductor is $12 \ mV$. The self-inductance of the inductor is . . . . . . $mH$.

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