Two inductor coils of self inductance, \( L_1 \) and \( L_2 \), having same length and same area of cross-section are placed so close that effective flux in one coil is completely linked with the other. The mutual inductance between them is equal to
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → \( \sqrt{L_1 L_2} \)
For two coils of the same length \( l \) and area \( A \), with \( N_1 \) and \( N_2 \) turns, the self inductances and the mutual inductance (all the flux of one linked with the other) are
\( \displaystyle L_1 = \frac{\mu_0 N_1^{2} A}{l}, \quad L_2 = \frac{\mu_0 N_2^{2} A}{l}, \quad M = \frac{\mu_0 N_1 N_2 A}{l} \)
Multiplying the two self inductances:
\( \displaystyle L_1 L_2 = \left(\frac{\mu_0 N_1 N_2 A}{l}\right)^{2} = M^{2} \)
\( \displaystyle M = \sqrt{L_1 L_2} \)
In general \( M = k\sqrt{L_1 L_2} \); complete flux linkage means the coupling coefficient \( k = 1 \).
Shortcut: Units decide it: \( M \) must be in henry, and only \( \sqrt{L_1 L_2} \) has the unit H (\( L_1 L_2 \) is \( \mathrm{H^2} \), \( L_1/L_2 \) has no unit and \( (L_1 L_2)^2 \) is \( \mathrm{H^4} \)).