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Q.Two coils A (primary) and B (secondary) have mutual inductance 2 x 10^-2 henry. If the current in the primary is i = 5 sin(10πt), then the maximum value of e.m.f. induced in coil B is

(a) π volt
(b) π/2 volt
(c) π/3 volt
(d) π/4 volt
Odisha ChseOdisha CHSE +2 Science Board Exam 2026MCQ· 1mImportance★★★★★
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Maximum induced emf = M×(di/dt)max=2×10−2×50π=πM\times(di/dt)_{max} = 2\times10^{-2}\times50\pi = \pi V.

The emf induced in the secondary coil due to mutual inductance MM is

e=−Mdidte = -M\dfrac{di}{dt}

Given i=5sin⁡(10πt)i = 5\sin(10\pi t), so

didt=5×10πcos⁡(10πt)=50πcos⁡(10πt)\dfrac{di}{dt} = 5\times10\pi\cos(10\pi t) = 50\pi\cos(10\pi t)

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