$A$ 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 $0.4 \, J T^{-1}$ is placed in a uniform magnetic field of $0.16 \, T$. The magnet is in stable equilibrium when the potential energy is ....... $J$.

Two magnetic dipoles $X$ and $Y$ are placed at a separation $d$,with their axes perpendicular to each other. The dipole moment of $Y$ is twice that of $X$. $A$ particle of charge $q$ is passing through their mid-point $P$,at an angle $\theta = 45^\circ$ with the horizontal line as shown in the figure. What would be the magnitude of the force on the particle at that instant? ($d$ is much larger than the dimensions of the dipole)

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 of $90^{\circ}$ from the magnetic meridian is $n$ times the corresponding work done to turn it through an angle of $60^{\circ}$. Find the value of $n$ (given $\cos 90^{\circ} = 0, \cos 60^{\circ} = 0.5$).

At a point on the equatorial line (right bisector) of a magnetic dipole,the magnetic: