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

Answer

C

Explanation

`epsi=-N(d phi)/(dt)`
` |(epsi)/(R)|=(N)/(R)(dphi)/(dt)`
`dq=(N)/(R) d phi`
`DeltaQ=(N(Deltaphi))/(R)`
`Delta Q= (Delta phi_("total"))/(R)`
=`((NBA))/(R)`
`=(mu_(0) ni pi r^(2))/(R)`
Putting values
`(4 pi xx 10^(-7) xx 100 xx 4xx piu xx (0.01)^(2))/(10 pi^(2))`
`DeltaQ=32 mu C`