When a current of 4 A through the primary coil gives rise to a total flux linkage of 1.35 Wb with the secondary coil, what is the coefficient of mutual induction (M)?
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → \( 0.34\,\mathrm{H} \)
A current \( I_1 \) in the primary coil sets up a total flux linkage \( N_2\Phi_2 \) with the secondary coil. The two are proportional, \( N_2\Phi_2=MI_1 \). The constant \( M \) is the coefficient of mutual induction, so it is the flux linkage per unit current in the primary.
\( \begin{aligned} M &= \frac{N_2\Phi_2}{I_1} \\ &= \frac{1.35\,\mathrm{Wb}}{4\,\mathrm{A}} \\ &= 0.3375\,\mathrm{H} \\ &\approx 0.34\,\mathrm{H} \end{aligned} \)
NCERT (2026-27):
Mutual Inductance – “We now consider the reverse case. A current I1 is passed through the solenoid S1 and the flux linkage with coil S2 is,
\( N_2\Phi_2=M_{21}I_1 \)\( \qquad (6.10) \)
M21 is called the mutual inductance of solenoid S2 with respect to solenoid S1.”