Two small magnets, each of magnetic moment $10 \, A \cdot m^2$, are placed in an end-on position at a distance of $0.1 \, m$ apart from their centers. The force acting between them is .... $N$.

$A$ bar magnet of magnetic moment $M$ is placed at a distance $D$ with its axis along the positive $X$-axis. Likewise, a second bar magnet of magnetic moment $M$ is placed at a distance $2D$ on the positive $Y$-axis and perpendicular to it as shown in the figure. The magnitude of the magnetic field at the origin is $|\vec{B}| = \alpha \left[ \frac{\mu_0}{4 \pi} \frac{M}{D^3} \right]$. The value of $\alpha$ must be (Assume $D \gg l$, where $l$ is the length of the magnets).

$A$ short bar magnet placed in a uniform magnetic field making an angle with the field experiences a torque. If the angle made by the magnet with the field is changed from $30^{\circ}$ to $45^{\circ}$, the torque of the magnet

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 potential at a point on the axial line of a bar magnet of dipole moment $M$ is $V$. What is the magnetic potential due to a bar magnet of dipole moment $\frac{M}{4}$ at the same point?

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