$A$ magnetic needle suspended parallel to a magnetic field requires $\sqrt{3} \text{ J}$ of work to turn it through $60^{\circ}$. The torque needed to maintain the needle in this position will be

$A$ magnet of magnetic moment $M$ is situated with its axis along the direction of a magnetic field of strength $B$. The work done in rotating it by an angle of $180^{\circ}$ will be

Two identical bar magnets of magnetic moment $M$ each,are placed along $X$ and $Y$-axes,respectively at a distance $d$ from the origin (as shown in the figure). The origin lies on the perpendicular bisector of the magnet placed on the $X$-axis and on the magnetic axis of the magnet placed on the $Y$-axis. If the magnitude of the total magnetic field at the origin is $B = \alpha \left[ \frac{\mu_0}{4 \pi} \frac{M}{d^3} \right]$,then the value of the constant $\alpha$ will be (given $d >> l$,where $l$ is the length of the bar magnets and the direction of $N$ to $S$ in the magnets is opposite with respect to each other).

$A$ small bar magnet has a magnetic moment $1.2 \, A-m^2$. The magnetic field at a distance $0.1 \, m$ on its axis will be: $(\mu_0 = 4\pi \times 10^{-7} \, T-m/A)$

$A$ bar magnet of magnetic moment $2 \text{ A m}^2$ lies aligned with the direction of a uniform magnetic field of $0.3 \text{ T}$. The amount of work required by an external torque to turn the magnet so as to align its magnetic moment normal to the field direction is (in $\text{ J}$)

$A$ 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