Due to a small magnet,the magnetic intensity at a distance $x$ in the end-on position (axial position) is $9 \ Gauss$. What will be the intensity at a distance $\frac{x}{2}$ in the broadside-on position (equatorial position)?

$A$ magnetic dipole of magnetic moment $ 6 \times 10^{-2} \text{ A m}^2 $ and moment of inertia $ 12 \times 10^{-6} \text{ kg m}^2 $ performs oscillations in a magnetic field of $ 2 \times 10^{-2} \text{ T} $. The time taken by the dipole to complete $ 20 $ oscillations is (assume $ \pi \simeq 3 $). (in $\text{ s}$)

Torques $\tau_1$ and $\tau_2$ are required for a magnetic needle to remain perpendicular to the magnetic fields of $B_1$ and $B_2$ at two different places. The ratio of $B_1: B_2$ is equal to

$A$ bar magnet of magnetic moment $1.5 \, A \cdot m^2$ lies aligned with the direction of a uniform magnetic field of $2 \, T$. What is the amount of work required to turn the magnet so as to align its magnetic moment normal to the field direction?

Two short bar magnets of magnetic moments '$M$' each are arranged at the opposite corners of a square of side '$d$',such that their centres coincide with the corners and their axes are parallel to one side of the square. If the like poles are in the same direction,the magnetic induction at any of the other corners of the square is

$A$ 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 $J \ T^{-1}$)