A long solenoid of radius 2 cm has 100 turns/cm and is surrounded by a 100- turn coil of radius 4cm having a total resistance of `20 Omega`. The coil is connected to a galvanometer as shown in fig. If the current in the solenoid is changed form 5 A in one direction to 5 A in the opposite direction, find the charge which flows through the galvanometer.

Explanation

If the current in the solenoid is I, the magnetic field inside the solenoid is ` B=mu parallel to its axis
Outside the soleniod, the field will be zero
The flux of the magnetic field through the coil will be `Phi =Bpir^2 N ` where r is the radius of the solenoid and N is the number of turns in the coil
The induced emf will have magnitude ` (d Phi)/(dt)= N pi r^2 (dB)/(dt) = pi r^2 N mu_0 n(du)/(dt)`
If R denotes the resistance of the coil, the current through the glavanometer is `I=(pir^2 N)/(R) mu_0 n (di)/(dt)` or, `I dt = (pir^2N)/(R) mu_0 n di`
The total charge passing through the galvanometer is ` DeltaQ = int i dt = (pi^2N)/(R) mu_0 n int di` = (pir^2 Nmu_0 n)/(R) Delta i ` =(pi(2 cm)^2 X 100 X 4 pi X 10^(-7) TmA^(-1) X100 cm^(-1) X 10 A)/(20 Omega) `~~ 8X 10^(-4) C = 800mu C