$A$ short bar magnet is placed in a uniform magnetic field of $2 \ T$ such that the axis of the magnet makes an angle of $45^{\circ}$ with the direction of the magnetic field. If the torque acting on the magnet is $0.36 \sqrt{2} \ Nm$,then the magnetic moment of the magnet is: (in $J \ T^{-1}$)

$A$ bar magnet of magnetic moment $M$ and moment of inertia $I$ is freely suspended such that the magnetic axial line is in the direction of the magnetic meridian. If the magnet is displaced by a very small angle $\theta$,the angular acceleration is (Magnetic induction of earth's horizontal field $= B_H$)

Two identical short bar magnets,each having a magnetic moment of $10 \, Am^2$,are arranged such that their axial lines are perpendicular to each other and their centres lie along the same straight line in a horizontal plane. If the distance between their centres is $0.2 \, m$,find the resultant magnetic induction at a point midway between them. (Given: $\mu_0 = 4\pi \times 10^{-7} \, Hm^{-1}$)

$A$ 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: $(\mu_0 = 4\pi \times 10^{-7} \, T-m/A)$

$A$ magnetic needle is placed in a uniform magnetic field and is aligned with the field. The needle is now rotated by an angle of $60^{\circ}$ and the work done is $W$. The torque on the magnetic needle at this position is

The magnetic potential at a point on the axial line of a bar magnet of dipole moment $M$ is $V$. What is the magnetic potential due to a bar magnet of dipole moment $\frac{M}{4}$ at the same point?