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(A) The self-induced $emf$ in an inductor is given by the formula $e = -L \frac{di}{dt}$.Since the inductors are identical,their self-inductance $L$ i

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(A) The self-induced $emf$ in an inductor is given by the formula $e = -L \frac{di}{dt}$.Since the inductors are identical,their self-inductance $L$ is the same.The magnitude of the induced $emf$ is $|e| = L \left| \frac{di}{dt} \right|$,where $\left| \frac{di}{dt} \right|$ represents the magnitude of the slope of the current-time graph.From the figure,the slope of line $1$ is steeper than the slope of line $2$,meaning $\left| \frac{di}{dt} \right|_1 > \left| \frac{di}{dt} \right|_2$.Therefore,the magnitude of the induced $emf$ is greater for inductor $1$,i.e.,$|e_1| > |e_2|$.

Two identical inductors carry currents that vary with time according to linear laws (as shown in the figure). In which of the two inductors is the self-induced $emf$ greater?

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