A long solenoid of diameter 0 . 1   m has 2 × 10 4 turns per meter. At the centre of the solenoid, a coil of 100 turns and radius 0 . 01   m is placed with its axis coinciding with the solenoid axis. The current in the solenoid reduces at a constant rate to 0   A from 4   A in 0 . 05   s . If the resistance of the coil is 10 π 2  Ω , the total charge flowing through the co

Step-by-step solution

In the given case, the solenoid produces a changing magnetic field. This field is linked with the coil inside and hence, there is an induced emf in the coil. We have, from the laws of EMI: ε =   - N d ϕ d t ε R = N R d ϕ d t The ratio given above is the induced current in the circuit which can further be written as the rate of flow of charge per unit time. d q = N R d ϕ Δ Q = N Δ ϕ R Δ Q = Δ ϕ t o t a l R = N B A R = N μ 0 n i π r 2 R Putting val