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: Check dimensions:
Mutual inductance has the dimension of $\mu_0\times\text{length}$. In the correct option, $\frac{r_2^2}{r_1}$ has dimension of length.

Step 2: Match:
Option (A) has $\frac{r_2}{r_1}$, which is dimensionless, and (B) and (C) have $r_2$ in the denominator, which makes M large for a tiny coil. Only (D), $\frac{\mu_0\pi r_2^2}{2r_1}$, fits the derivation above.

Final Answer:
$\frac{\mu_0\pi r_2^2}{2r_1}$. \[ \boxed{M = \frac{\mu_0 \pi r_2^2}{2r_1}} \]