A compass needle free to turn in a horizontal plane is placed at the centre of 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 0.35 A, the needle points west to east.
(a) Determine the horizontal component of the 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.
(a) Number of turns in the coil (n) = 30
Radius of coil (r) = 12cm = 0.12m
Current in the coil (I) = 0.35A
Angle of dip (δ) = 45o
Magnetic fields due to current carrying coils (B) = 
where, μ0 = Permeability of free space = 4π × 10-7TmA1
B = 
= 5.49 × 10-5 T
∴ Horizontal component of the earth’s magnetic field = B sinδ
= 5.49 × 10-5 T × sin45o
= 5.49 × 10-5 T × 0.707
= 3.88 × 10-5 T
(b) When 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, then the needle will reverse its original direction (i.e. the needle will point from East to West).