If a bar magnet of moment $10^{-4} Am^2$ is kept in a uniform magnetic field of $12 \times 10^{-3} T$ such that it makes an angle of $30^{\circ}$ with the direction of the magnetic field, then the torque acting on the magnet is:

Two short bar magnets have their magnetic moments $1.2 \text{ A m}^2$ and $1.0 \text{ A m}^2$. They are placed on a horizontal table parallel to each other at a distance of $20 \text{ cm}$ between their centres,such that their north poles point towards geographic south. They have a common magnetic equatorial line. The horizontal component of the Earth's magnetic field is $3.6 \times 10^{-5} \text{ T}$. The resultant horizontal magnetic induction at the midpoint of the line joining their centers is $\left(\frac{\mu_0}{4 \pi} = 10^{-7} \text{ N/A}^2\right)$.

The work done in rotating a bar magnet from its equilibrium position by $60^{\circ}$ in a magnetic field is $W$. What is the torque required to hold it in this position?

$A$ short bar magnet produces a magnetic field of $6.4 \times 10^{-5} \,T$ at a distance of $20 \,cm$ from the centre of the magnet on the normal bisector of the magnet. The magnetic field produced by this magnet at a distance of $40 \,cm$ from the centre of the magnet on the axis,is

$A$ small bar magnet has a magnetic moment $1.2 \, A-m^2$. The magnetic field at a distance $0.1 \, m$ on its axis will be: $(\mu_0 = 4\pi \times 10^{-7} \, T-m/A)$

$A$ magnet of magnetic moment $20$ $C.G.S.$ units is freely suspended in a uniform magnetic field of intensity $0.3$ $C.G.S.$ units. The amount of work done in deflecting it by an angle of $30^{\circ}$ in $C.G.S.$ units is: