Question

Two coils having self-inductance \(L_1 = 75\) mH and \(L_2 = 48\) mH are coupled with each other. If the mutual inductance of the coils is 37.2 mH, then coefficient of coupling will be

• A. \(0.58\)
• B. \(0.60\)
• C. \(0.62\)
• D. \(0.64\)

Hint:

Find the flux of the large coil field through the small coil, treating the field as uniform.

Answer 1

The Correct Option is C

Step 1: Field at the centre:
The big coil (radius \(r_1\), one turn) carries a current I and produces a field at the centre \(B = \frac{\mu_0I}{2r_1}\).

Step 2: Flux through the small coil:
Since \(r_2 \ll r_1\), the field is nearly uniform over the small coil: \(\Phi = B\cdot\pi r_2^2 = \frac{\mu_0I\,\pi r_2^2}{2r_1}\).

Step 3: Mutual inductance:
\[ M = \frac\Phi I = \frac{\mu_0\pi r_2^2}{2r_1} \]

Final Answer:
The mutual inductance is \(\frac{\mu_0\pi r_2^2}{2r_1}\), option (D). \[ \boxed{M = \frac{\mu_0 \pi r_2^2}{2r_1}} \]