If a current change of 6 A in one coil causes the flux in the second coil of 2000 turns to change by $12 × 10^{-4} Wb$ per turn, the mutual inductance of the pair of coils would be

0.4 H

The correct answer is Option (2) → 0.4 H

Mutual inductance $M$ is given by:

$M = \frac{N \, \Delta \Phi}{\Delta I}$

Given: $N = 2000$ turns, $\Delta \Phi_\text{per turn} = 12 \times 10^{-4} \, \text{Wb}$, $\Delta I = 6 \, \text{A}$

Total flux linkage:

$N \Delta \Phi = 2000 \cdot 12 \times 10^{-4} = 2.4 \, \text{Wb-turns}$

Mutual inductance:

$M = \frac{2.4}{6} = 0.4 \, \text{H}$

Mutual inductance M = 0.4 H