# A short bar magnet of magnetic moment $m= 0.32 \;J/T$. is placed in a uniform magnetic field of $0.15 \;T$. If the bar is free to rotate in the plane of the field, Which orientation would correspond to its unstable equilibrium ? What is the potential energy of the magnet?

$\begin{array}{1 1} -4.8 \times 10^{-2}\;J \\ 4.8 \times 10^{-2}\;J \\ 48 \times 10^{-2}\;J \\ -48 \times 10^{-2}\;J \end{array}$

Solution :
Given :
$m= 0.32 \;J/T$
$B= 0.15 \;T$
$\theta= 180^{\circ}$
For unstable equilibrium . the angle between the magnetic moment and the magnetic field is $180^{\circ}$
the potential energy of the magnet.
$U= -mB \cos \theta^{\circ}$
$\qquad= 0.32 \times 0.15 \times (-1)$
$(\cos 180^{\circ} =-1)$
$\qquad= =4.8 \times 10^{-2}\;J$
Answer : $4.8 \times 10^{-2}\;J$