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

$A$ bar magnet of magnetic moment $M$ and moment of inertia $I$ is freely suspended such that the magnetic axial line is in the direction of the magnetic meridian. If the magnet is displaced by a very small angle $\theta$, the angular acceleration is (Magnetic induction of earth's horizontal field $= B_H$)

$A$ magnet of magnetic moment $M$ is rotated through $360^{\circ}$ in a magnetic field $H$,the work done will be

Two equal bar magnets are kept as shown in the figure. The direction of the resultant magnetic field,indicated by the arrowhead at point $P$,is (approximately):

Two identical magnetic dipoles of magnetic moments $1.0 \, A-m^2$ each are placed at a separation of $2 \, m$ with their axes perpendicular to each other. The resultant magnetic field at a point $P$ midway between the dipoles is:

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: