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

$A$ 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 $= B_H$)

$A$ short bar magnet of magnetic moment $0.21 \ A \cdot m^2$ is placed with its axis perpendicular to the direction of the horizontal component of the earth's magnetic field. The distance of the point on the axis of the magnet from the centre of the magnet where the resultant magnetic field is inclined at $45^{\circ}$ with the horizontal component of the earth's field direction is (horizontal component of the earth's magnetic field $= 4.2 \times 10^{-5} \ T$). (in $cm$)

$A$ magnetic dipole of magnetic moment $M$ is freely suspended in a magnetic field of induction $B$. The minimum and maximum values of the potential energy of the dipole,respectively,are

Write an expression for the magnitude of the magnetic field at a point lying on the equatorial line of a bar magnet.