Two different coils have self-inductance 8 m | vedclass

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(A) The induced $emf$ $(e)$ in a coil is given by the formula: $e = L \frac{di}{dt}$.Here,$L_1 = 8 \ mH$ and $L_2 = 2 \ mH$.The rate of change of curr

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$4: 1$

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(A) The induced $emf$ $(e)$ in a coil is given by the formula: $e = L \frac{di}{dt}$.Here,$L_1 = 8 \ mH$ and $L_2 = 2 \ mH$.The rate of change of current,$\frac{di}{dt}$,is the same for both coils.Therefore,the ratio of the induced $emf$s is:$\frac{e_1}{e_2} = \frac{L_1 (di/dt)}{L_2 (di/dt)} = \frac{L_1}{L_2}$.Substituting the given values:$\frac{e_1}{e_2} = \frac{8 \ mH}{2 \ mH} = \frac{4}{1}$.Thus,the ratio is $4: 1$.

Two different coils have self-inductance $8 \ mH$ and $2 \ mH$. The current in both coils is increased at the same constant rate. The ratio of the induced $emf$s in the coils is

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