Write the equation of torque acting on a bar magnet placed in a uniform magnetic field.

$A$ 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:

$A$ short bar magnet of magnetic moment $M=0.32 \; J \, T^{-1}$ is placed in a uniform magnetic field of $0.15 \; T$. If the bar is free to rotate in the plane of the field,which orientation would correspond to its $(a)$ stable,and $(b)$ unstable equilibrium? What is the potential energy of the magnet in each case?

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 $\text{ T}$)

$A$ magnetic dipole is under the influence of two orthogonal magnetic fields,$B_1 = 0.5 \times 10^{-3} \ T$ and $B_2 = 0.866 \times 10^{-3} \ T$. If the dipole comes to stable equilibrium at an angle $\theta$ with respect to the $B_2$ field,then the value of $\theta$ is (in $^{\circ}$)

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