The work done in rotating a bar magnet,which is initially in the direction of a uniform magnetic field,through $45^{\circ}$ is $W$. The additional work to be done to rotate the magnet further through $15^{\circ}$ is

If there is no torsion in the suspension thread, then the time period of a magnet executing $SHM$ is

The magnetic field due to a short magnet at a point on its axis at a distance $X \, cm$ from the center of the magnet is $200 \, G$. The magnetic field at a point on the equatorial axis at a distance $X \, cm$ from the center of the magnet is.....$G$.

Two magnets $A$ and $B$ are identical and are arranged as shown in the figure. Their length is negligible in comparison to the separation between them. $A$ magnetic needle is placed between the magnets at point $P$,which gets deflected through an angle $\theta$ under the influence of the magnets. The ratio of distances $d_1$ and $d_2$ will be:

The magnitude of the axial field due to a bar magnet at a distance of $1 \ m$ is found to be $5 \times 10^{-8} \ T$. The magnetic moment of the bar magnet is $\left(\mu_0 = 4 \pi \times 10^{-7} \ T \ m/A\right)$. (in $A \ m^2$)

Write the equation of torque acting on a bar magnet placed in a uniform magnetic field.