Ans:
Given,
Mutual inductance µ = 1.5H
Current at initial point is given by I1= 0 A
Current at final point is given by I2 = 20 A
Therefore, change in current becomes
dI = I1- I2 = = 20 – 0 = 20 A
Time taken for the change is given by t = 0.5s
We know that the expression for induced emf is –
$e=\frac{d\phi }{dt}$
Where, Phi represents the change in Induced flux
Now, the relation between the emf and inductance is given by –
\[e=\mu \frac{dI}{dt}\]
Upon equating both the above equations, we get –
\[\frac{d\phi }{dt}=\mu \frac{dI}{dt}\]
\[\frac{d\phi }{dt}=1.5\times 20=30Wb\]
Therefore, the change in flux linkage is 30 Wb.