Magnetic intensity for an axial point due to a short bar magnet of magnetic moment $M$ is given by

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

$A$ magnet of total magnetic moment $10^{-2} \hat{i} \text{ A} \cdot \text{m}^2$ is placed in a time-varying magnetic field $\vec{B} = B \cos(\omega t) \hat{i}$, where $B = 1 \text{ T}$ and $\omega = 0.125 \text{ rad/s}$. The work done to reverse the direction of the magnetic moment at $t = 1 \text{ s}$ is: (in $\text{ J}$)

The magnetic moment of a magnet is $2.5 \, J \, T^{-1}$. How much work in $J$ must be done to rotate it from its stable equilibrium position to an unstable equilibrium position in a magnetic field of $0.2 \, T$?

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: $\frac{\mu_0}{4 \pi} = 10^{-7} \text{ N/A}^2$)

In which direction would the magnetic field on the axis at a distance $Z$ from the centre of the bar magnet be?