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

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):

Two identical short bar magnets,each having magnetic moment $M$,are placed a distance of $2d$ apart with axes perpendicular to each other in a horizontal plane. The magnetic induction at a point midway between them is

$A$ bar magnet of magnetic moment $M$ and moment of inertia $I$ is freely suspended such that the magnetic axial line is in the direction of the magnetic meridian. If the magnet is displaced by a very small angle $\theta$, the angular acceleration is (Magnetic induction of earth's horizontal field $= B_H$)

$A$ magnet of magnetic moment $20$ $C.G.S.$ units is freely suspended in a uniform magnetic field of intensity $0.3$ $C.G.S.$ units. The amount of work done in deflecting it by an angle of $30^{\circ}$ in $C.G.S.$ units is:

$A$ short bar magnet of magnetic moment $0.21 \ A \cdot m^2$ is placed with its axis perpendicular to the direction of the horizontal component of the earth's magnetic field. The distance of the point on the axis of the magnet from the centre of the magnet where the resultant magnetic field is inclined at $45^{\circ}$ with the horizontal component of the earth's field direction is (horizontal component of the earth's magnetic field $= 4.2 \times 10^{-5} \ T$). (in $cm$)