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 = 3.0X 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.