Question

A magnetic needle of pole strength $20\times {10}^{-4}$ Am is pivoted at it's centre. Its $N-pole$ is pulled eastward by a string. The horizontal force required to produce a deflection of ${30}^o$ from the magnetic meridian is (take $B_H={10}^{-4}T)$ is :

• A. $4\times {10}^{-7}N$
• B. $2\times {10}^{-7}N$
• C. $\frac{2}{\sqrt{3}}\times {10}^{-7}N$
• D. $4\sqrt{3}\times {10}^{-7}N$

Answer 1

The correct option is C $\frac{2}{\sqrt{3}}\times {10}^{-7}N$

Consider $\Delta ABC$

Here, $\overrightarrow{AC}=\overrightarrow{F_H}=HorizontalForce$

$\overrightarrow{BC}=\overrightarrow{F_M}=MagneticForce=m\times B_H=20\times {10}^{-4}\times {10}^{-4}=2\times {10}^{-7}$

$\overrightarrow{AB}==ResultantForce$

So, $\tan {60}^0=\frac{F_M}{F_H}=\frac{2\times {10}^{-7}}{F_H}$

$\Rightarrow F_H=\frac{2\times {10}^{-7}}{\sqrt{3}}$

Therefore, Horizontal Force required to produce a deflection of

30o30o from the magnetic meridian is $\frac{2\times {10}^{-7}}{\sqrt{3}}$