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 $^{\circ}$)

$A$ small bar magnet placed with its axis at $30^{\circ}$ with an external field of $0.06\, T$ experiences a torque of $0.018\, Nm$. The minimum work required to rotate it from its stable to unstable equilibrium position is

$A$ short bar magnet placed with its axis at $30^{\circ}$ with a uniform external magnetic field of $0.25 \ T$ experiences a torque of magnitude $4.5 \times 10^{-2} \ J$. The magnitude of the magnetic moment of the magnet will be . . . . . . $J \ T^{-1}$.

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 $60^{\circ}$. The value of $n$ is:

$A$ bar magnet of magnetic moment $10^4\,J/T$ is free to rotate in a horizontal plane. The work done in rotating the magnet slowly from a direction parallel to a horizontal magnetic field of $4 \times 10^{-5}\,T$ to a direction $60^\circ$ from the field will be... $J$.

$A$ short magnetic needle is placed in a magnetic field $B \hat{i}$. The needle is oriented in the direction $(\sqrt{3} \hat{i} + \hat{j})$. The needle experiences a torque of $0.06 \ N-m$. If the same magnetic needle is placed in a magnetic field $2B \hat{j}$ and is oriented in the direction $(\hat{i} + \sqrt{3} \hat{j})$,the torque experienced by it is: (in $N-m$)