$A$ 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.....$J$

The dipole moment of a short bar magnet is $1.25 \, A-m^2$. The magnetic field on its axis at a distance of $0.5 \, m$ from the centre of the magnet is:

Two equal bar magnets are kept as shown in the figure. The direction of the resultant magnetic field,indicated by the arrowhead at point $P$,is (approximately):

$A$ bar magnet of length $6\,cm$ has a magnetic moment of $4\,J\,T^{-1}$. Find the strength of magnetic field at a distance of $200\,cm$ from the centre of the magnet along its equatorial line.

Two identical small bar magnets, each of dipole moment $3\sqrt{5} \text{ J/T}$, are placed at a center-to-center separation of $10 \text{ cm}$, with their axes perpendicular to each other as shown in the figure. The value of the magnetic field at the point $P$ midway between the magnets is $\alpha \times 10^{-3} \text{ T}$. The value of $\alpha$ is . . . . . . .

Two short bar magnets $A$ and $B$ (having magnetic moments $M_{1}$ and $M_{2}$ respectively) are kept one above the other with their magnetic axes perpendicular to each other. If their resultant magnetic field at a point on the axis of magnet $A$ is inclined at $45^{\circ}$ with the axis of magnet $A$,then the ratio of magnetic moments $\frac{M_{2}}{M_{1}}$ is $[\tan 45^{\circ} = 1]$.