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, N = 30
Current in the coil, I = 0.35 A
Radius of the circular coil, r = 12 cm = 0.12 m. 
The magnetic field at the centre of the coil is $\mathrm{B} = \frac{{\mathrm{\mu }}_0 \mathrm{NI}}{2\mathrm{r}}.$

It acts in a direction perpendicular to the plane of the coil.
Its component parallel to the magnetic meridian is $\frac{{\mathrm{\mu }}_0\mathrm{NI}}{2\mathrm{r}} \cos 45°.$
The component perpendicular to the magnetic meridian is $\frac{{\mathrm{\mu }}_0\mathrm{NI}}{2\mathrm{r}}\sin 45°$.

Given that the needle points in the west-east direction. 
Horizontal component of earth's magnetic field is given by 
${\mathrm{B}}_{\mathrm{H}} = \frac{{\mathrm{\mu }}_0 \mathrm{NI}}{2\mathrm{r}} \cos 45°$
    $$\begin{gathered} = \frac{4\mathrm{\pi } \times {10}^{-7} \times 30 \times 0.35}{2 \times 0.12 \times \sqrt{2}} \\ = 0.39 \times {10}^{-4}\mathrm{T} \end{gathered}$$

${\mathrm{B}}_{\mathrm{H}} = 0.39 \mathrm{G}$

(b) When the current in the coil is reversed and coil is turned through 90^o anticlockwise, the plane of the coil makes an angle of 45° with the magnetic meridian on the other side. Thus, needle will rotate and will set in east to west direction.