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

Two short bar magnets of magnetic moments $M$ and $\sqrt{3} M$ are joined like a cross (+). This cross is suspended from its centre with its plane horizontal in the earth's magnetic field. When the cross comes to equilibrium,the angle made by the weaker magnet with the earth's magnetic field $B_H$ is (in $^{\circ}$)

The work done in turning a magnet of magnetic moment $M$ by an angle $\theta$ from its initial position parallel to the magnetic meridian is:

Two identical small bar magnets, each of dipole moment $3\sqrt{5} \text{ J/T}$, are placed at a center-to-center separation of $10 \text{ cm}$, with their axes perpendicular to each other as shown in the figure. The value of the magnetic field at the point $P$ midway between the magnets is $\alpha \times 10^{-3} \text{ T}$. The value of $\alpha$ 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... $J$.

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)?