A long solenoid of radius 2 cm has 100 turns/cm and carries a current of 5A. A coil of radius 1 cm having 100 turns and a total resistance of 20 Ω is placed inside the solenoid coaxially. Thecoil is connected to a galvanometer. If the current in the solenoid is reversed in direction, find the charge that flows through the galvanometer.

 

 r = 2 cm = 2 x 10^–2 m n = 100 turns / cm = 10000 turns/m i = 5 A B = μ0 ni = 4π x 10^–7 x 10000 x 5 = 20π x 10^–3 = 62.8 x 10^–3 T n base 2 = 100 turns R = 20 Ω r = 1 cm = 10^–2 m Flux linking per turn of the second coil = Bπ r^2 = Bπ x 10^–4 ϕ base 1 = Total flux linking = Bn base 2 π r2 = 100 x π x 10^–4 x 20π x 10^–3 When current is reversed. ϕ base 2 = – ϕ base 1 dϕ = ϕ base 2 – ϕ base 1 = 2 x 100 x π x 10^–4 x 20π x 10–3 E = dϕ/dt = 4π^2 x 10^-4/dt l = E/R = 4π^2 x 10^-4/dt x 20 q = idt = 4π^2 x 10^-4/dt x 20 x dt = 2 x 10^-4 C. 

 

Regards

Arun (askIITians forum expert)