Question

A bar magnet of magnetic moment $2.0A-m^2$ is free to rotate about a vertical axis through its centre. The magnet is released from rest from the east-west position. The kinetic energy of the magnet as it takes the north-south position is found to be $\alpha \times {10}^{-8}J$, where $\alpha$ and $\beta$ are the single digit integers. Then find the value of $\alpha /\beta$. The horizontal component of the earth's magnetic field is $B=25\mu T$.Earth's magnetic field is from south to north.

Answer 1

Potential energy of bar magnet is given by

$U=-\overrightarrow{M}\ast \overrightarrow{B}$

initial angle between $\overrightarrow{M}$ and $\overrightarrow{B}={90}^{\circ }$

Final angle between $\overrightarrow{M}$ and $\overrightarrow{B}=0^{\circ }$

$\therefore U_i=-MBcos{90}^{\circ }=0$

$U_f=-MBcos{0}^{\circ }=-MB$

$\therefore \Delta U=U_f-U_i=-MB$

and hence work done $=-MB$

$=-2\ast 25\ast {10}^{-6}$

$-5\ast {10}^{-5}J$

$\therefore$ KE gained by magnet

$=5\ast {10}^{-5}J$

in question KE=$\alpha \ast {10}^{-\beta }$

$\therefore \alpha =5,\beta =5$

$\therefore \frac{\alpha }{\beta }=1$