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° 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.

Answer 1

(a) given,number of turns , n = 30,

radius of circular coil, r = 12cm = 12 × 10^-2 m

and current through the coil, i = 0.35A. we have to find horizontal component of earth's magnetic field, H

As it is clear from figure shown the needle can point west to east only when

H = Bsin45°

where B = magnetic field strength due to current in coil = $\frac{\mu_0}{4\pi}\frac{2\pi ni}{r}$

$\therefore\:H=\frac{\mu_0}{4\pi}\frac{2\pi ni}{r}sin45^{\circ}\\\\=10^{-7}\times\frac{2\pi\times30\times0.35}{12\times10^{-2}}.\frac{1}{\sqrt{2}}\\\\=2\times\frac{22}{7}\times\frac{30\times35}{12\times\sqrt{2}}\times10^{-7}=3.9\times10^{-5}T$

(b) when current in coil is reversed and oil is turned through 90° anticlockwise, the direction of needle will reverse (e.g., it will point from east to west ).this follows from figure shown.

Answer 2

Answer:

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 45o with the magnetic meridian. When the current in the coil is 0.35 A, the needle points west to east.

Top answer

(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.

A compass needle free to turn in a horizontal plane