Due to a small magnet,the magnetic intensity at a distance $x$ in the end-on position (axial position) is $9 \ Gauss$. What will be the intensity at a distance $\frac{x}{2}$ in the broadside-on position (equatorial position)?

The work done in rotating a bar magnet of magnetic moment $M$ from its unstable equilibrium position to its stable equilibrium position in a uniform magnetic field $B$ is .........

$A$ bar magnet of magnetic moment $10^4\,J/T$ is free to rotate in a horizontal plane. The work done in rotating the magnet slowly from a direction parallel to a horizontal magnetic field of $4 \times 10^{-5}\,T$ to a direction $60^\circ$ from the field will be... $J$.

Two short bar magnets, each with a magnetic moment of $9 \text{ Am}^2$, are placed such that one is at $x = -3 \text{ cm}$ and the other is at $y = -3 \text{ cm}$. If their magnetic moments are directed along the positive and negative $X$-directions respectively, then the resultant magnetic field at the origin is: (in $\text{ T}$)

Write the equation of potential energy of a bar magnet placed in a uniform magnetic field.

What is the magnitude of the equatorial magnetic field due to a bar magnet of length $5.0 \ cm$ at a distance $75 \ cm$ from its midpoint? The magnetic moment of the bar magnet is $0.75 \ A \ m^2$.