$A$ 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

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

$A$ 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 $\text{ s}$)

Two short bar magnets have their magnetic moments $1.2 \text{ A m}^2$ and $1.0 \text{ A m}^2$. They are placed on a horizontal table parallel to each other at a distance of $20 \text{ cm}$ between their centres,such that their north poles point towards geographic south. They have a common magnetic equatorial line. The horizontal component of the Earth's magnetic field is $3.6 \times 10^{-5} \text{ T}$. The resultant horizontal magnetic induction at the midpoint of the line joining their centers is $\left(\frac{\mu_0}{4 \pi} = 10^{-7} \text{ N/A}^2\right)$.

The magnetic field due to a small magnetic dipole of magnetic moment $M$,at a distance $r$ from the centre on the equatorial line is given by (in $M.K.S.$ system):

$A$ 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 .... $N-m$.