$A$ short bar magnet of magnetic moment $0.4\,JT^{-1}$ is placed in a uniform magnetic field of $0.16\,T$. The magnet is in stable equilibrium when the potential energy is

$A$ magnet is parallel to a uniform magnetic field. If it is rotated by $60^o$,the work done is $0.8\, J$. How much work is done in moving it $30^o$ further (in $, J$)?

$A$ short bar magnet of magnetic moment $M=0.32 \; J \, T^{-1}$ 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 $(a)$ stable,and $(b)$ unstable equilibrium? What is the potential energy of the magnet in each case?

$A$ magnetic needle lying parallel to a magnetic field requires $W$ units of work to turn it through $60^{\circ}$. The torque required to maintain the needle in this position will be

$A$ small bar magnet of magnetic moment $M$ is placed in a uniform magnetic field $H$. If the magnet makes an angle of $30^{\circ}$ with the field,the torque acting on the magnet is:

Write the equation of potential energy of a bar magnet placed in a uniform magnetic field.