Question

A bar magnet of magnetic moment 3.0 A-m${}^2$ is free to rotate about a vertical axis passing through its centre. The magnet is released from rest from east - west position. Then the kinetic energy of the magnet as it takes North- South position is :
(horizontal component of earths magnetic field is 25$\mu$T)

• A. 25 $\mu$ J
• B. 75 $\mu$J
• C. 100$\mu$ J
• D. 12.5 $\mu$ J

Answer 1

The correct option is B 75 $\mu$J

$m=3A{m}^2$
$B=25\mu T$
$\Delta K.E.=\Delta P.E.$

$P=-m\cdot B$ $=-mBcos\theta$
${\theta }_1=90$
${\theta }_2=0$
Kinetic energy $=P_2-P_1$
$=mB\cos {\theta }_1-mB\cos {\theta }_2$
$=3\times 25\mu$
$\therefore$ Kinetic energy of the Bar magnet will be 75 $\mu J$