A long solenoid of diameter `0.1 m` has `2 xx 10^(4)` turns per metre. At the centre of the solenoid, a `100-`turns coil of radius `0.01 m` is placed with its axis coinciding with the constant rate from `+ 2 A` to `2 A` in `0.05 s`. Find the total charge (in `mu C`) flowing through the coil during this time when the resistance of the coil is `40 !=^(2) Omega`.

Answer

`(8)`

Explanation

The mutual inductance of solenoid coil system
`M = mu_(0)n_(1)N_(2)A_(2) = mu_(0)n_(1)N_(2)pir_(2)^(2)`
`= 4pi xx 10^(-7) xx 2xx 10^(4) xx 100 xx pi xx (0.01)^(2)`
`= 8pi^(2) xx 10^(-5) H`
`EMF` induced in the coil: `e_(2) = - M(Deltai)/(Deltat)`
`= - 8pi^(2) xx 10^(-5) xx ((-2 - 2)/(0.005)) = 640 pi^(2) xx 10^(-5) V` Required charge:
`q = i Deltat = (e)/(R ) Deltat = (640 xx pi^(2) xx 10^(-5))/(40 pi^(2)) xx 0.05 = 8 muC`