Question

A bar magnet of length $16cm$ has a pole strength of $500\times {10}^{-3}Am$ The angle at which it should be placed to the direction of external magnetic field of induction $2.5G$ so that it may experience a torque of $\sqrt{3}\times {10}^{-5}$ Nm is :

• A. $\pi$
• B. $\frac{\pi }{2}$
• C. $\frac{\pi }{3}$
• D. $\frac{\pi }{6}$

Answer 1

The correct option is C $\frac{\pi }{3}$

$l=16m$
$P=500\times {10}^{-3}Am$
$z=\overrightarrow{m}\times \overrightarrow{B}$
$B=2.5G=2.5\times {10}^{-4}T$
$m=lP$
torque due to magnetic field $z=\sqrt{3}\times {10}^{-5}N.m$
$z=mBm\times sinQ$
$\sqrt{3}\times {10}^{-5}=lPBsinQ$
$\sqrt{3}\times 16\times 500\times 2.5\times sinQ.$
$SinQ=\frac{\sqrt{3}}{2}$
$Q=\pi /3$