Question

A magnetic dipole of moment $2.5\, Am ^{2}$ 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 $=2.5\, Am ^{2}$
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 =M B \sin \theta $
$KE =$work done $=W=\int\limits_{\theta_{1}}^{\theta_{2}} \tau \cdot d \theta$
$=M B\left(\cos \theta_{1}-\cos \theta_{2}\right)$

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