$A$ bar magnet of magnetic moment $1.5 \, J/T$ lies aligned with the direction of a uniform magnetic field of $0.22 \, T$. What is the amount of work required by an external torque to turn the magnet so as to align its magnetic moment perpendicular to the field direction? (in $J$)

The effect due to a uniform magnetic field on a freely suspended magnetic needle is as follows:

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

$A$ magnetic dipole is under the influence of two magnetic fields. The angle between the field directions is $60^{\circ},$ and one of the fields has a magnitude of $1.2 \times 10^{-2}\;T$. If the dipole comes to stable equilibrium at an angle of $15^{\circ}$ with this field,what is the magnitude of the other field?

The rate of change of torque $\tau$ with respect to deflection $\theta$ is maximum for a magnet suspended freely in a uniform magnetic field of induction $B$ when $\theta = ........ ^\circ$.

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: