The Total solution for NCERT class 6-12
A compass needle free toturn in a horizontal plane is placed at the centre of circular coil of 30 turnsand radius 12 cm. The coil is in a vertical plane making an angle of 45ºwith the magnetic meridian. When the current in the coil is 0.35 A, the needlepoints west to east.
(a) Determine thehorizontal component of the earth’s magnetic field at the location.
(b) The current in thecoil is reversed, and the coil is rotated about its vertical axis by an angleof 90º in the anticlockwise sense looking from above. Predict the direction ofthe needle. Take the magnetic declination at the places to be zero.
Number of turns in thecircular coil, N = 30
Radius of the circularcoil, r = 12 cm = 0.12 m
Current in thecoil, I = 0.35 A
Angle of dip, δ =45°
Where,
= Permeability offree space = 4π × 10−7 Tm A−1
= 5.49 × 10−5 T
The compass needlepoints from West to East. Hence, the horizontal component of earth’s magneticfield is given as:
BH = Bsin δ
= 5.49 × 10−5 sin45° = 3.88 × 10−5 T = 0.388 G
(b) When the current inthe coil is reversed and the coil is rotated about its vertical axis by anangle of 90 º, the needle will reverse its original direction. In this case,the needle will point from East to West.