$A$ magnetic needle of magnetic moment $6 \times 10^{-2} \text{ A m}^2$ and moment of inertia $9.6 \times 10^{-5} \text{ kg m}^2$ performs simple harmonic motion in a magnetic field of $0.01 \text{ T}$. The time taken to complete $10$ oscillations is (Take $\pi = 3.14$): (in $\text{ s}$)

The ratio of magnetic fields due to a bar magnet at two axial points $P_1$ and $P_2$,which are separated from each other by $10 \ cm$,is $25: 2$. Point $P_1$ is situated at $10 \ cm$ from the center of the magnet. (Points $P_1$ and $P_2$ are on the same side of the magnet,and the distances of $P_1$ and $P_2$ from the center are much greater than the distance of the two ends of the magnet from the center). The magnetic length of the bar magnet is: (in $cm$)

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

Points $A$ and $B$ are situated perpendicular to the axis of a small bar magnet at large distances $x$ and $3x$ from its centre on opposite sides. The ratio of the magnetic fields at $A$ and $B$ will be approximately equal to

The magnetic moment of a bar magnet is $0.5 \text{ A m}^2$. It is suspended in a uniform magnetic field of $8 \times 10^{-2} \text{ T}$. The work done in rotating it from its most stable to most unstable position is:

$A$ magnetic dipole experiences a torque of $80 \sqrt{3} \ N \ m$ when placed in a uniform magnetic field in such a way that the dipole moment makes an angle of $60^{\circ}$ with the magnetic field. The potential energy of the dipole is