Question

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 ${45}^0$ with the magnetic meridian when the current in the coil is $0.35$A, the needle points west to east. Determine the horizontal component of earth's magnetic field at the location.

• A. $3.9\times {10}^{-7}$ tesla
• B. $3.9\times {10}^{-5}$ tesla
• C. $8.0\times {10}^{-5}$ tesla
• D. $7.0\times {10}^{-7}$ tesla

Answer 1

The correct option is B $3.9\times {10}^{-5}$ tesla

Here, $n=30,r=12cm=12\times {10}^{-2}m.i=0.35A,H=?$

As it is clear from figure shown the needle can point west to east only when $H=Bsin{45}^0$

where, B = magnetic field strength due to current in coil $\frac{{\mu }_0}{4\pi }\frac{2\pi ni}{r}$

$\therefore H=\frac{{\mu }_0}{4\pi }\frac{2\pi ni}{r}sin{45}^0={10}^{-7}\times \frac{2\pi \times 30\times 0.35}{12\times {10}^{-2}}.\frac{1}{\sqrt{2}}=2\times \frac{22}{7}\times \frac{30\times 35}{12\times \sqrt{2}}\times {10}^{-7}=3.9\times {10}^{-5}$ tesla