Question

A bar magnet of magnetic moment $2A{m}^2$ is free to rotate about a vertical axis passing through its center. The magnet is released from rest from east - west position. Then the KE of the magnet as it takes N-S position is
$(B_H=25\mu T)$

• A. $25\mu J$
• B. $50\mu J$
• C. $100\mu J$
• D. $12.5\mu J$

Answer 1

The correct option is B $50\mu J$

$m=2A{m}^2$
$B_H=25\mu T$
${\theta }_1=\pi /2$
${\theta }_2=0$
${\mu }_1=-mB\cos \pi /2$
$=0$
${\mu }_2=-mB\cos 0^o$
$=-2\times 25\mu T$
$k_1=0$
$k_2-k_1=U_1-U_2$
$k_2=50\mu J$