$A$ magnetic dipole in a constant magnetic field has

$A$ bar magnet has a magnetic moment of $200 \text{ A m}^2$. The magnet is suspended in a magnetic field of $0.30 \text{ N A}^{-1} \text{ m}^{-1}$. The torque required to rotate the magnet from its equilibrium position through an angle of $30^{\circ}$ will be:

The magnitude of the axial field due to a short bar magnet at a distance of $50 \,cm$ from its mid-point is (The magnetic moment of the bar magnet is $0.4 \,Am^2$)

$A$ bar magnet having a magnetic moment of $2 \times 10^4 \, JT^{-1}$ is free to rotate in a horizontal plane. $A$ horizontal magnetic field $B = 6 \times 10^{-4} \, T$ exists in the space. The work done in taking the magnet slowly from a direction parallel to the field to a direction $60^{\circ}$ from the field is.....$J$

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:

$A$ bar magnet of magnetic moment $200 \, A-m^2$ is suspended in a magnetic field of intensity $0.25 \, N/A-m$. The torque required to deflect it through $30^\circ$ is .... $N-m$.