Question

A compass needle is free to turn in a horizontal plane is placed at the centre of a circular coil of $30$ turns and radius $12\text{cm}$. The coil is in a vertical plane, making an angle of ${45}^{\circ }$ with the magnetic meridian. When the current in the coil is $0.35\text{A}$, the needle points west to east. Find the horizontal component of the earth's magnetic field at the location.

• A. $0.38\text{G}$
• B. $0.19\text{G}$
• C. $0.24\text{G}$
• D. $0.09\text{G}$

Answer 1

The correct option is A $0.38\text{G}$

Given data for coil:
$\text{Number of turns,}N=30\text{Radius},r=12\text{cm}=0.12\text{m}\text{Current through the coil,}I=0.35\text{A}\text{and Angle of dip,}\delta ={45}^{\circ }$

The magnetic field at the centre of the coil having $N$ turns is,

$B=\frac{{\mu }_{0}NI}{2r}$

$\therefore B=\frac{4\pi \times {10}^{-7}\times 30\times 0.35}{2\times 0.12}$

$=5.49\times {10}^{-5}\text{T}$( directed along the axis of the coil.)


As the compass needle points from west to east. Thus, the horizontal component of earth's magnetic field balances the field produced by the coil,

$\therefore B_H=B\cos \delta$

$=5.49\times {10}^{-5}\times \frac{1}{\sqrt{2}}\text{T}$

$=3.88\times {10}^{-5}\text{T}=0.38\text{G}$

<!--td {border: 1px solid #ccc;}br {mso-data-placement:same-cell;}--> Hence, $\left(A\right)$ is the correct answer.