A coil of radius 10 cm and resistance `40 Omega` has 1000 turns. It is placed with its plane vertical and its axis parallel to the magnetic meridian. The coil is connected to a galvanometer and is rotated about the vertical diameter through an angle of `180^@`. Find the charge which flows through the galvanometer if the horizontal component of the earth's magnetic field is `B_H = 3.0xx 10^(-5) T`.

Explanation

Here `r=10cm=0.1m,R=40 Omega,`
`N=1000`
`theta=180^(@), B_H=3xx10^(-5)T.`
`phi=N(B.A)=NBA cos 180^(@)`
`=-NBA`
`=-1000xx3xx10^(-5)pixx 1xx1xx10^(-2)`
`=3 pixx10^(-4) weber`
` d phi= 2NBA= 6picc10^(-4) weber`
`e=(d phi)/(dt)=(6 pi xx10^(-4)V)/(dt)`
`and i- (6 pi xx10^(-4))/(40dt)=(4.71xx10^(-5))/(dt)`
`Q=(4.71xx10^(-5)xxdt)/(dt)`
`=4.71xx10^(-5)C.