Question

A magnetic needle of pole strength $20\sqrt{3}\text{Am}$ is pivoted at its center. Its $N$-pole is pulled eastward by a string. The horizontal force required to produced a deflection of ${30}^{\circ }$ from magnetic meridian $[\text{Take}{B}_H={10}^{-4}\text{T}]$ is

• A. $4\sqrt{3}\times {10}^{-3}\text{N}$
• B. $\frac{2}{\sqrt{3}}\times {10}^{-3}\text{N}$
• C. $2\times {10}^{-3}\text{N}$
• D. $4\times {10}^{-3}\text{N}$

Answer 1

The correct option is D $4\times {10}^{-3}\text{N}$

Given,$m=20\sqrt{3}\text{Am};B_H={10}^{-4}\text{T};\theta ={30}^{\circ }$


Deflection torque = Restoring torque

$\text{F}\times \text{perpendicular distance}$ = $M{B}_H\sin \theta$

$\Rightarrow F\times OA=M{B}_H\sin \theta ......\left(1\right)$

From $△OAF,$

$\cos \theta =\frac{OA}{ON}$

$\Rightarrow \left(ON\right)\cos \theta =OA=l\cos \theta$

From, (1) we get,

$F=\frac{M{B}_H\sin \theta }{\left(OA\right)}$

$=\frac{m2l{B}_H\sin \theta }{l\cos \theta }$

$=2m{B}_H\tan \theta$

$=2\times 20\sqrt{3}\times {10}^{-4}\times \frac{1}{\sqrt{3}}[\because \theta ={30}^{\circ }]$

$=4\times {10}^{-3}\text{N}$

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