The magnetic moment of a bar magnet is $0.5 \text{ A m}^2$. It is suspended in a uniform magnetic field of $8 \times 10^{-2} \text{ T}$. The work done in rotating it from its most stable to most unstable position is:

The magnetic moment of a magnet is $2.5 \, J \, T^{-1}$. How much work in $J$ must be done to rotate it from its stable equilibrium position to an unstable equilibrium position in a magnetic field of $0.2 \, T$?

$A$ short bar magnet of magnetic moment $0.21 \ A \cdot m^2$ is placed with its axis perpendicular to the direction of the horizontal component of the earth's magnetic field. The distance of the point on the axis of the magnet from the centre of the magnet where the resultant magnetic field is inclined at $45^{\circ}$ with the horizontal component of the earth's field direction is (horizontal component of the earth's magnetic field $= 4.2 \times 10^{-5} \ T$). (in $cm$)

Figure shows a small magnetised needle $P$ placed at a point $O$. The arrow shows the direction of its magnetic moment. The other arrows show different positions (and orientations of the magnetic moment) of another identical magnetised needle $Q$.
$(a)$ In which configuration the system is not in equilibrium?
$(b)$ In which configuration is the system in $(i)$ stable,and $(ii)$ unstable equilibrium?
$(c)$ Which configuration corresponds to the lowest potential energy among all the configurations shown?

The work done in turning a magnet of magnetic moment $M$ by an angle of $90^{\circ}$ from the magnetic meridian is $n$ times the corresponding work done to turn it through an angle of $60^{\circ}$. The value of $n$ is:

Two short bar magnets of magnetic moments '$M$' each are arranged at the opposite corners of a square of side '$d$',such that their centres coincide with the corners and their axes are parallel to one side of the square. If the like poles are in the same direction,the magnetic induction at any of the other corners of the square is