$A$ magnetic dipole is under the influence of two magnetic fields. The angle between the field directions is $60^{\circ},$ and one of the fields has a magnitude of $1.2 \times 10^{-2}\;T$. If the dipole comes to stable equilibrium at an angle of $15^{\circ}$ with this field,what is the magnitude of the other field?

The magnetic moment of a magnet is $20 \, C.G.S$ units. How much work is required to rotate it by $30^{\circ}$ from its equilibrium position in a magnetic field of $0.3 \, C.G.S$ units?

$A$ bar magnet of length $10 \text{ cm}$ and having the pole strength equal to $10^{-3} \text{ A-m}$ is kept in a magnetic field having magnetic induction $B$ equal to $4 \pi \times 10^{-3} \text{ T}$. It makes an angle of $30^{\circ}$ with the direction of magnetic induction. The value of the torque acting on the magnet is

Two small bar magnets are placed in a line with like poles facing each other at a certain distance $d$ apart. If the length of each magnet is negligible as compared to $d$,the force between them will be inversely proportional to

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).

$A$ magnet of magnetic moment $2 \,J \,T^{-1}$ is aligned in the direction of a magnetic field of $0.1 \,T$. What is the net work done to bring the magnet normal to the magnetic field (in $\,J$)?