The work done in rotating a magnet of magnetic moment $2 \, A-m^2$ in a magnetic field of $5 \times 10^{-3} \, T$ from the direction along the magnetic field to the opposite direction to the magnetic field is:

$A$ bar magnet is held perpendicular to a uniform magnetic field. The couple acting on the magnet is to be halved by rotating it. Through what angle should it be rotated?

Two short bar magnets of magnetic moments '$M$' each are arranged at the opposite corners of a square of side '$d$',such that their centres coincide with the corners and their axes are parallel to one side of the square. If the like poles are in the same direction,the magnetic induction at any of the other corners of the square is

Give the stability position of a bar magnet for $\theta = 0^{\circ}$ and $\theta = 180^{\circ}$.

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

The pole strength of a bar magnet is $48 \, A \cdot m$ and the distance between its poles is $25 \, cm$. The torque required to hold the magnet at an angle of $30^{\circ}$ with a uniform magnetic field of flux density $0.15 \, N \cdot A^{-1} \cdot m^{-1}$ is . . . . . . $N \cdot m$.