Question

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 ${45}^o$ 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 ${90}^o$ in the anticlockwise sense looking from above. Predict the direction of the needle. Take the magnetic declination at the places to be zero.

Answer 1

(a)

The needle will point along W-E if the result of the earth’s magnetic field and magnetic field due to the coil have a resultant in the W-E direction. This happens if $B\cos {45}^{\circ }$=Earth's field

$⟹\frac{{\mu }_{0}nI}{2r}cos{45}^{\circ }=\frac{4\pi \times {10}^{-7}\times 30\times 0.35}{2\times 0.12}\times \frac{1}{\sqrt{2}}=0.39G$=Earth's magnetic field.

(b)

Rotation of coil about its axis has no effect on the magnetic field because of symmetry about the axis. As the direction of current is reversed, so will be the direction of its field. Component of magnetic field along west to east will reverse in direction i.e. it will become east to west.

Component of field of coil along North to south will also reverse to south to north. Hence, total field along S-N axis is not zero anymore.

Hence Net field is from South-East to North-West.