$A$ compass needle free to turn in a horizontal plane is placed at the centre of a circular coil of $30$ turns and radius $12 \;cm$. The coil is in a vertical plane making an angle of $45^{\circ}$ 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 $90^{\circ}$ in the anticlockwise sense looking from above. Predict the direction of the needle. Take the magnetic declination at the place to be zero.

(A) Given:
Number of turns,$N = 30$
Radius,$r = 12 \;cm = 0.12 \;m$
Current,$I = 0.35 \;A$
Angle between coil plane and magnetic meridian,$\theta = 45^{\circ}$
$(a)$ The magnetic field $B$ at the centre of the coil is $B = \frac{\mu_0 N I}{2r} = \frac{4\pi \times 10^{-7} \times 30 \times 0.35}{2 \times 0.12} = 5.497 \times 10^{-5} \;T$.
The component of $B$ perpendicular to the magnetic meridian is $B \sin(45^{\circ})$. Since the needle points west to east,this component must balance the horizontal component of the Earth's magnetic field $B_H$.
$B_H = B \sin(45^{\circ}) = 5.497 \times 10^{-5} \times \frac{1}{\sqrt{2}} \approx 3.89 \times 10^{-5} \;T = 0.389 \;G$.
$(b)$ Reversing the current reverses the direction of $B$. Rotating the coil by $90^{\circ}$ anticlockwise changes the orientation of the coil plane relative to the meridian. The net effect of these operations results in the magnetic field of the coil still opposing the Earth's horizontal component in the same line of action. Thus,the needle will point from East to West.