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JEE PHYSICS MCQ: ELECTRO MAGNETIC INDUCTION

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Application +4 | -1
The current in a coil changes from 0 to 2 A2 \mathrm{~A} in 0.05 s0.05 \mathrm{~s}. If the induced emf is 80 V80 \mathrm{~V}, the self-inductance of the coil is,
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Correct option: D
2 H2 \mathrm{~H}
The correct answer is 2 H2 \mathrm{~H} because it is calculated using the formula for self-inductance, L=εdIdtL = \frac{\varepsilon}{\frac{dI}{dt}}.

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