A pair of adjacent coils has a mutual inductance of 1.5 H.

Solution:

Mutual inductance of a pair of coils, µ = 1.5 H

Initial current, I1 = 0 A

Final current I2 = 20 A

Change in current, $d I=I_{2}-I_{1}=20-0=20 \mathrm{~A}$

Time taken for the change, $t=0.5 \mathrm{~s}$

Induced emf, $e=\frac{d \phi}{d t}$    ...(1)

Where $d \phi$ is the change in the flux linkage with the coil.

Emf is related with mutual inductance as:

$e=\mu \frac{d I}{d t}$   ...(2)

Equating equations (1) and (2), we get

$\frac{d \phi}{d t}=\mu \frac{d I}{d t}$

$d \phi=1.5 \times(20)$

$=30 \mathrm{~Wb}$

Hence, the change in the flux linkage is 30 Wb.