Write an expression for the magnitude of the magnetic field at a point lying on the equatorial line of a bar magnet.

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

Points $A$ and $B$ are situated perpendicular to the axis of a $2 \ cm$ long bar magnet at large distances $X$ and $3X$ from its centre on opposite sides. The ratio of the magnetic fields at $A$ and $B$ will be approximately equal to:

Two short magnets of equal dipole moments $M$ are fastened perpendicularly at their centres, which lie at the origin. Let the two magnets lie along the $X$-axis and $Y$-axis, respectively. The magnitude of the magnetic field at a distance $R$ from the centre on the $Y$-axis is $\frac{\mu_0}{4 \pi} \frac{M_0}{R^3}$. Assuming $R >> l$ (magnet length), the magnitude of $M$ is

$A$ short bar magnet of magnetic moment $0.4 \, J T^{-1}$ is placed in a uniform magnetic field of $0.16 \, T$. The magnet is in stable equilibrium when the potential energy 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: $\frac{\mu_0}{4 \pi} = 10^{-7} \text{ N/A}^2$)