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JEE PHYSICS MCQ: ELECTRO MAGNETIC INDUCTION

1 Medium
1 of 1
Application +4 | -1
Self-inductance of two coils is 2H&8H2 \mathrm{H} \& 8 \mathrm{H}. If 50%50 \% flux of primary coil is linked with secondary coil, then find coefficient of mutual inductance.
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Correct option: A
2 H2 \mathrm{~H}
Option A (2 H) is correct because the mutual inductance is calculated using the formula M = k*sqrt(L1*L2), where k is the coupling coefficient.

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