Question

A short bar magnet of moment $4{\text{Am}}^2$ is placed in the magnetic merdian with its south-pole pointing geographic north. The distance between two null points is found to be $20\text{cm}$.

$\left(i\right)$The value of horizontal component of the earth's magnetic field at the place is

• A. $4\times {10}^{-4}\text{T}$
• B. $6\times {10}^{-4}\text{T}$
• C. $8\times {10}^{-4}\text{T}$
• D. $10\times {10}^{-4}\text{T}$

Answer 1

The correct option is C $8\times {10}^{-4}\text{T}$

Representation of the data given in the question is given in the figure below:

$M=4{\text{Am}}^2;d=10\text{cm}$

At null point : $B_{ax}=B_H$

$\therefore \frac{{\mu }_0}{4\pi }\frac{2M}{d^3}=B_H$

Substituting the data gives,

$\frac{4\pi \times {10}^{-7}}{4\pi }\times \frac{2\times 4}{{10}^{-3}}=B_H$

$\Rightarrow B_H=8\times {10}^{-4}\text{T}$