The coefficient of mutual inductance of two circuits `A` and `B` is `3 mH` and their respective resistances are `10 Omega` and `4 Omega`. How much current should change in `0.02 s` in circuit `A`, so that the induced current in `B` should be `0.0060 A`?

Options

A. `0.24 A`
B. `1.6 A`
C. `0.18 A`
D. `0.16 A`

Explanation

Induced current in `B = 0.006 A = 6 xx 10^(-3) A`
Induced emf in `B = 6 xx 10^(-3) xx 4V = 24 xx 10^(-3) V`
Now, `M(dI)/(dt) = 24 xx 10^(-3)`
or `dI = (24 xx 10^(-3) xx 0.02)/(3 xx 10^(-3))A = 0.16 A`