$A$ small bar magnet has a magnetic moment $1.2 \, A-m^2$. The magnetic field at a distance $0.1 \, m$ on its axis will be: $(\mu_0 = 4\pi \times 10^{-7} \, T-m/A)$

$A$ 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. If the like poles are in the same direction,the magnetic induction at any of the other corners of the square is

$A$ magnetic dipole of magnetic moment $M$ is freely suspended in a magnetic field of induction $B$. The minimum and maximum values of the potential energy of the dipole,respectively,are

$A$ bar magnet having a magnetic moment of $2.0 \times 10^{5} \; J T^{-1}$ is placed along the direction of a uniform magnetic field of magnitude $B = 14 \times 10^{-5} \; T$. The work done in rotating the magnet slowly through $60^{\circ}$ from the direction of the field is .............. $J$.

Two short bar magnets have magnetic moments $1.2 \text{ Am}^2$ and $1.0 \text{ Am}^2$. 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 the geographic south. They share a common magnetic equatorial line. The horizontal component of the Earth's magnetic field is $3.6 \times 10^{-5} \text{ T}$. Calculate the resultant horizontal magnetic induction at the midpoint of the line joining their centres. (Given: $\frac{\mu_0}{4 \pi} = 10^{-7} \text{ N/A}^2$)