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 ${45}^o$ 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, $n=30,r=12$cm $=12\times {10}^{-2}$m
$i=0.35$A, H$=?$
As it is clear from figure shown the needle can point west to east only when $H=B\sin {45}^o$
where, B$=$magnetic field strength due to current in coil $=\frac{{\mu }_o}{4\pi }\frac{2\pi ni}{r}$
$\therefore H=\frac{{\mu }_o}{4\pi }\frac{2\pi ni}{r}\sin {45}^o$
$={10}^{-7}\times \frac{2\pi \times 30\times 0.35}{12\times {10}^{-2}}\cdot \frac{1}{\sqrt{2}}$
$=2\times \frac{22}{7}\times \frac{30\times 35}{12\times \sqrt{2}}\times {10}^{-7}$
$=3.9\times {10}^{-5}$ tesla.