A long solenoid of diameter 0.1 m has `2 xx 10^(4)` turns per meter. At centre of the solenoid is 100 turns coil of radius 0.01 m placed with its axis coinciding with solenoid axis. The current in the solenoid reduce at a constant rate to 0A from 4 a in 0.05 s . If the resistance of the coil is `10 pi^(2) Omega`, the total charge flowing through the coil during this time is

Options

A. `16 mu C`
B. `32 mu C `
C. `16pi mu C `
D. `32 pi mu C `

Answer

B

Explanation

Given `n = 2 xx 10^4 , I = 4A`
Initial I = 0A
`therefore B_i = 0 " or " phi_i = 0`
Finally , the magnetic field at the centre of the solenoid is given as
`B_f = mu_0 nI = 4pi xx 10^(-7) xx 2 xx 10^4 xx 4 = 32 pi xx 10^(-3) T`
Finl magnetic flux through the coil is given as
`phi_f = NBA = 100 xx 32 pi xx 10^(-3) xx pi xx (0.01)^2`
`phi_f = 32 pi^2 xx 10^(-5) Tm^2`
Induced charge `q = (|Delta phi|)/( R) = (|phi_f - phi_i|)/( R) = (32 pi^2 xx 10^(-5))/(10 pi^2)`
` = 32 xx 10^(-6) C = 32 mu C`