A compass needle free to turn in a horizontal plane is placed at the centre of a circular coil of 30 turns and radius 12 cm. The coil is in a vertical plane making an angle of `45^@` with the magnetic meridian when the current in the coil is `0.35amp.`, the needle points west to east.
(a) Determine the horizontal component of earth's magnetic field at the location.
(b) The current in the coil is reversed and the coil is rotated about its vertical axis by an angle of `90^@` in the anticlockwise sense looking from above. Predict the direction of the needle. Take the magnetic declination at the places to be zero.

Explanation

(a) The magnetic field at the centre of the coil of radius r, number of turns N and carrying current I is given by:
` B = (mu_0 IN)/(2r)`
Component of magnetic field parallel to meridian
` = (mu_0 IN)/(2r) cos 45^@`
The component of magnetic field in the magnetic meridian should be equal and opposite to that of the coil:
`-(mu_0IN)/(2r) cos 45^@ + H = 0`
` rArr H = (mu_0 IN)/(2r) cos 45^@`
Here, N = 30, r = 12 cm = 0.12 m, I = 0.35 A
` rArr H = (4pi xx 10^(-7) xx 30 xx 0.35)/(2 xx.12) xx (1)/(sqrt2)`
`= 0.39 xx 10^(-4) T = 0.39 G`
(b) In this case, plane of the coil makes an angle of `45^@` with the magnetic meridian on the other side. The needle will rotate and set in the east-to-west direction.