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Question

# Two coils are at fixed locations. When coil-1 has no current and current in the coil-2 increases at the rate 15.0 As−1. The e.m.f. in the coil-1 is 25.0 mV. When coil-2 has no current and coil-1 has a current of 3.6 A., the flux linkage in coil-2 is-

A
16 mWb
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B
10 mWb
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C
4 mWb
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D
6 mWb
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Solution

## The correct option is D 6 mWbCase-(I) When current in the coil-1 is zero and current in the coil-2 increases at a rate of (di2dt)=15 As−1 and emf in the coil-1 is E1=25 mV. Using, the principle of mutual inductance, ϕ1=Mi2 On differentiating both sides we get, |E1|=dϕ1dt=Mdi2dt .......(1) Where, ϕ1=Flux linked with the coil-1 M=Co-efficient of mutual inductancei2=Current in the coil-2 Putting the values in (1) we get, 25×10−3=M×15 ⇒M=53×10−3 H Case-(II) Now, flux linked with coil-2 when current in coil-1 is 3.6 A is, =Mi1 =53×10−3×3.6 =6×10−3 Wb=6 mWb <!--td {border: 1px solid #ccc;}br {mso-data-placement:same-cell;}--> Hence, (D) is the correct answer. Why this question ? This question tests your basic understanding of mutual inductance.

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