Question

A bar magnet of magnetic moment $2{\text{Am}}^2$ is free to rotate about a vertical axis passing through its centre. The magnet is then released from rest at east -west position. Find the Kinetic energy of the magnet as it takes north-south position is ________$\mu \text{J}$ $(B_H=25\mu \text{T})$ (Give integer value)

Answer 1

Kinetic energy of barmagnet in the presence of external magnetic field is , $K=M{B}_H[1-\cos \theta ]$

From the data given in the question, $\theta ={90}^{\circ }\Rightarrow \cos {90}^{\circ }=0$

$\therefore K=M{B}_H$

Substituting the given data we get,

$K=2\times 25\times {10}^{-6}$
$=50\times {10}^{-6}\text{J}$
$=50\mu \text{J}$

Accepted answer : $50$