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:

Write the equation of potential energy of a bar magnet placed in a uniform magnetic field.

$A$ short bar magnet of magnetic moment $2.5 \text{ Am}^2$ is kept in a uniform magnetic field of $4 \times 10^{-5} \text{ T}$. The work done in moving the magnet from its most stable position to its most unstable position is:

$A$ bar magnet of length $3 \, cm$ has points $A$ and $B$ along its axis at distances of $24 \, cm$ and $48 \, cm$ on the opposite sides. The ratio of the magnetic fields at these points is

$A$ torque of $1.732 \times 10^{-5} \text{ Nm}$ is required to hold a magnet at $90^{\circ}$ with the horizontal component of the Earth's magnetic field. The torque required to hold it at $60^{\circ}$ will be $\left[\sin 90^{\circ}=1, \sin 60^{\circ}=\frac{\sqrt{3}}{2}\right] [\sqrt{3} = 1.732]$

$A$ bar magnet is held perpendicular to a uniform magnetic field. If the couple acting on the magnet is to be halved by rotating it,then the angle by which it is to be rotated is....$^o$