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.

Solution:

Number of turns in the circular coil, N = 30

Radius of the circular coil, r = 12 cm = 0.12 m

Current in the coil, I = 0.35 A

Angle of dip, δ = 45°

(a) The magnetic field due to current I, at a distance r, is given as:

$B=\frac{\mu_{0} 2 \pi N I}{4 \pi r}$

Where,

$\mu_{0}=$ Permeability of free space $=4 \pi \times 10^{-7} \mathrm{Tm} \mathrm{A}^{-1}$

$\therefore B=\frac{4 \pi \times 10^{-7} \times 2 \pi \times 30 \times 0.35}{4 \pi \times 0.12}$

$=5.49 \times 10^{-5} \mathrm{~T}$

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

BH = Bsin δ

= 5.49 × 10−5 sin 45° = 3.88 × 10−5 T = 0.388 G

 

(b) When the current in the coil is reversed and the coil is rotated about its vertical axis by an angle of 90 º, the needle will reverse its original direction. In this case, the needle will point from East to West.