$A$ short bar magnet placed with its axis at $30^{\circ}$ with an external field of $800 G$ experiences a torque of $0.016 Nm$. The magnetic moment of the bar magnet is (in $Am^2$)

The dipole moment of a short bar magnet is $1.25 \, A-m^2$. The magnetic field on its axis at a distance of $0.5 \, m$ from the centre of the magnet is:

Two identical short bar magnets are placed at $120^{\circ}$ as shown in the figure. The magnetic moment of each magnet is $M$. Then the magnetic field at the point $P$ on the angle bisector is given by

$A$ magnet of magnetic moment $2 \, J \, T^{-1}$ is aligned in the direction of a magnetic field of $0.1 \, T$. What is the net work done to bring the magnet normal to the magnetic field?

$A$ 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:

$A$ short bar magnet of magnetic moment $5.25 \times 10^{-2} \;J\, T^{-1}$ is placed with its axis perpendicular to the earth's field direction. At what distance from the centre of the magnet,the resultant field is inclined at $45^{\circ}$ with the earth's field on
$(a)$ its normal bisector and
$(b)$ its axis.
Magnitude of the earth's field at the place is given to be $0.42 \;G$. Ignore the length of the magnet in comparison to the distances involved.