A long solenoid with a radius of 2 cm has 100 turns/cm and carries a 5 A current. Coaxially positioned inside the solenoid is a coil of radius 1 cm , consisting of 100 turns and a total resistance of 20 . This coil is connected to a galvanometer. Determine the charge flown through the galvanometer if the direction of the current in the solenoid is reversed.

Step-by-step solution

The magnetic field inside the long solenoid is given by B = ₀ n I Substituting the given values, n = 100 turns/cm = 10^4 turns/m and I = 5 A : B = 4 10⁻⁷ 10^4 5 = 2 10⁻² T The initial magnetic flux linked with the inner coil is _i = N B A Substituting N = 100 and A = r^2 = (10⁻²)^2 = 10⁻⁴ m ^2 : _i = 100 (2 10⁻²) ( 10⁻⁴) = 2 ^2 10⁻⁴ Wb When the current in the solenoid is reversed, the final magnetic flux becomes _f = -2 ^2 10⁻⁴ Wb The magnitude of the change in flux is | | = | _f - _i| = 4 ^2 10⁻⁴ Wb The charge flo