Question

A compass needle free to turn in a horizontal plane is placed at thecentre of circular coil of 30 turns and radius 12 cm. The coil is in avertical plane making an angle of 45° with the magnetic meridian.When the current in the coil is 0.35 A, the needle points west toeast.(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 aboutits vertical axis by an angle of 90° in the anticlockwise senselooking from above. Predict the direction of the needle. Take themagnetic declination at the places to be zero.

Answer 1

Given: Number of turns in the coil is 30, its radius is 12 cm, the coil is in a vertical plane making an angle of 45º with the magnetic meridian and the current in the coil is 0.35 A.

a)

The magnetic field at a distance from the coil is given as,

B= μ 0 nI 2πr

Where, n is the number of turns in the coil, I is the current in the wire and r is the radius of the coil.

By substituting the given values in the above expression, we get

B= 4π× 10 −7 ×30×0.35 2π×0.12 =5.49× 10 −4  T

The compass needle points from West to East. Hence, the horizontal component of earth’s magnetic field is given as,

B h =Bsinδ

Where, δ is the angle of dip.

By substituting the given values in the above expression, we get

B h =5.49× 10 −5 sin45 =3.88× 10 −5 =0.39 G

Thus the horizontal component of the earth’s magnetic field is 0.39 G.

b)

When the current in the coil is reversed and the coil is rotated about its vertical axis by an angle of 90°, the direction of the needle will get reversed. Thus, the needle will point from East to West.