Question

A bar magnet of magnetic moment $M$ is divided into '$n$' equal parts by cutting parallel to length. Then one part is suspended in a uniform magnetic field of strength $2T$ and held making an angle ${60}^0$ with the direction of the field. When the magnet is released, the kinetic energy of the magnet in the equilibrium position is:

• A. $\frac{M}{n}$J
• B. $\frac{M}{n^2}$J
• C. $Mn$ J
• D. $M{n}^2$ J

Answer 1

Answer is A
Net magnetic moment after cutting$=\frac{M}{n}$
We know that, $|U|=MB\cos \theta$
In equilibrium position, all the potential energy will be converted into kinetic energy i.e. $KE=|U|$
Given $M_{new}=\frac{M}{n},B=2T,\theta ={60}^{\circ }$
Therefore, $K.E=\frac{M}{n}\times 2\times \cos {60}^{\circ }=\frac{M}{n}J$