The midpoints of two small magnetic dipoles of length $d$ in end-on positions are separated by a distance $x$ $(x >> d)$. The force between them is proportional to $x^{-n}$,where $n$ is:

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 .... $N-m$.

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 $A \ m^2$)

$A$ magnet of magnetic moment $M$ is rotated through $360^{\circ}$ in a magnetic field $H$. The work done will be:

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