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 $20\text{cm}$.

$\left(ii\right)$If the magnet is turned through ${90}^{\circ }$ , the resultant field at the location of the null point will be.

• A. $2\sqrt{5}\times {10}^{-4}\text{T}$
• B. $4\sqrt{5}\times {10}^{-4}\text{T}$
• C. $3\sqrt{5}\times {10}^{-4}\text{T}$
• D. $8\sqrt{5}\times {10}^{-4}\text{T}$

Answer 1

The correct option is B $4\sqrt{5}\times {10}^{-4}\text{T}$


$B_R=\sqrt{B_H^2+B_{eq}^2}=\sqrt{B_H^2+{\left(\frac{B_{ax}}{2}\right)}^2}$

$\Rightarrow B_R=\sqrt{B_H^2+{\left(\frac{B_H}{2}\right)}^2}=B_H\sqrt{1+\frac{1}{4}}$

$\Rightarrow B_R=\frac{\sqrt{5}}{2}{B}_H$

Substituting the value gives,

$B_R=\frac{\sqrt{5}}{2}\times 8\times {10}^{-4}=4\sqrt{5}\times {10}^{-4}\text{T}$