Question

A magnetic needle of pole strength $20\sqrt{3}$ Am is pivoted at its centre.Its N -pole is pulled eastward by a string.The horizontal force required to produce a deflection of ${30}^o$ from magnetic meridian (taken $B_H={10}^{-4}T$) is :

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

Answer 1

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

Given,

$B_H={10}^{-4}T$

$m=20\sqrt{3}Am$

$\theta ={30}^0$ deflection from the magnetic meridian

The horizontal required force ,

$F=mB$. . . . . .. . . . . .(1)

we know that, the magnetic meridian $B_H$

$B_H=Bcos\theta$

$B=\frac{B_H}{cos\theta }=\frac{{10}^{-4}}{cos{30}^0}$

$B=1.1547\times {10}^{-4}T$

From equation (1),

$F=20\sqrt{3}\times 1.1547\times {10}^{-4}N$

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

The correct option is A.