Question

A magnetic dipole of moment $2.5A{m}^2$ is free to rotate about a vertical axis passing through its centre. It is released from East-West direction. Its kinetic energy at the moment, it takes North-South position is
$\left(B_H=3\times 10^{-5}T\right)$

• A. $50\mu J$
• B. $100\mu J$
• C. $175\mu J$
• D. $75\mu J$

Answer 1

Answer: (D)

Given, dipole moment of magnet $=2.5A{m}^2$
Earth's magnetic field, $B_H=3\times 10^{-5}T$
When the magnetic dipole is released from East-West direction, a torque acts on it. So, in the displacement from $E-W$ to $N-S$, work is done by the torque. The value of torque
$\tau =M\times B=MB\sin \theta$
$KE=$work done $=W=\underset{{\theta }_1}{\overset{{\theta }_2}{\int }}\tau \cdot d\theta$
$=MB\left(\cos {\theta }_1-\cos {\theta }_2\right)$

$\therefore KE=MB\cos 0^{\circ }$
$=2.5\times 3\times 10^{-5}$
$=7.5\times 10^{-5}J=75\times 10^{-6}\mu J$