$A$ magnetic dipole is acted upon by two magnetic fields which are inclined to each other at an angle of $75^{\circ}$. One of the fields has a magnitude of $15 \, mT$. The dipole attains stable equilibrium at an angle of $30^{\circ}$ with this field. The magnitude of the other field (in $mT$) is close to:

$A$ magnet is parallel to a uniform magnetic field. If it is rotated by $60^\circ$,the work done is $0.8\, J$. How much work is done in moving it $30^\circ$ further?

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 .........

The magnetic field due to a short magnet at a point on its axis at a distance $X \, cm$ from the center of the magnet is $200 \, G$. The magnetic field at a point on the equatorial axis at a distance $X \, cm$ from the center of the magnet is.....$G$.

Two magnets $A$ and $B$ are identical and are arranged as shown in the figure. Their length is negligible in comparison to the separation between them. $A$ magnetic needle is placed between the magnets at point $P$,which gets deflected through an angle $\theta$ under the influence of the magnets. The ratio of distances $d_1$ and $d_2$ will be:

$A$ bar magnet of magnetic moment $\vec{M}$ is placed in a magnetic field of induction $\vec{B}$. The torque exerted on it is . . . . . . .