The magnetic moment of a magnet is $20 \, C.G.S$ units. How much work is required to rotate it by $30^{\circ}$ from its equilibrium position in a magnetic field of $0.3 \, C.G.S$ units?

$A$ bar magnet of length $6\,cm$ has a magnetic moment of $4\,J\,T^{-1}$. Find the strength of magnetic field at a distance of $200\,cm$ from the centre of the magnet along its equatorial line.

$A$ 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

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:

$A$ donut-shaped permanent magnet (magnetization parallel to the axis) can slide frictionlessly on a vertical rod. Treat the magnets as dipoles with mass $m_d$ and dipole moment $M$. When we put two back-to-back magnets on the rod,the upper one will float. At what height $z$ does it float?

Two small magnets, each of magnetic moment $10 \, A \cdot m^2$, are placed in an end-on position at a distance of $0.1 \, m$ apart from their centers. The force acting between them is .... $N$.