Question

A bar magnet of moment $2$ $A-m^2$ is free to rotate about a Vertical axis passing through its centre. The magnet is released form rest from east west direction. The $K.E.$ of the magnet as it takes north-south direction is ($B_H=$25$X{10}^{-6}T$)

• A. $25$ x ${10}^{-6}$ J
• B. $50$ x ${10}^{-6}$ J
• C. $75$ x ${10}^{-6}$ J
• D. $100$ x ${10}^{-6}$ J

Answer 1

Answer: (B)

$50$ x ${10}^{-6}$ J

Given:

Magnetic moment, m= 2Am${}^2$

B${}_H=25\times {10}^{-6}T$

${\theta }_1=\frac{\Pi }{2}$

$\tgeta_2 = 0

Solution:

$U_1=-mBCos{\theta }_1$

$\Rightarrow U_1=-2\times 25\times {10}^{-6}Cos\frac{\Pi }{2}$

$\Rightarrow U_1=0$

$U_2=-mBCos{\theta }_2$

$\Rightarrow U_2=-2\times 25\times {10}^{-6}\times Cos0$

$\Rightarrow U_2=-50\times {10}^{-6}$

Change in K.E. = $U_1-U_2=0-(-50\times {10}^{-6})$

$\Rightarrow ChangeinK.E.=50\times {10}^{-6}J$

Hence, option B is correct.