$A$ bar magnet has a magnetic moment of $200 \text{ A m}^2$. The magnet is suspended in a magnetic field of $0.30 \text{ N A}^{-1} \text{ m}^{-1}$. The torque required to rotate the magnet from its equilibrium position through an angle of $30^{\circ}$ will be:

$A$ 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 $cm$)

$A$ magnet of length $0.1 \, m$ and pole strength $10^{-4} \, A \cdot m$ is kept in a magnetic field of $30 \, Wb/m^2$ at an angle of $30^\circ$. The couple (torque) acting on it is $...... \times 10^{-4} \, N \cdot m$.

$A$ short bar magnet placed with its axis at $30^{\circ}$ with an external field of $800 G$ experiences a torque of $0.016 Nm$. The magnetic moment of the bar magnet is (in $Am^2$)

Two short bar magnets of magnetic moments $m$ each are arranged at the opposite corners of a square of side $d$ such that their centres coincide with the corners and their axes are parallel. If the like poles are in the same direction,the magnetic induction at any of the other corners of the square is

$A$ magnet of magnetic moment $4 \, A-m^2$ is held in a uniform magnetic field $5 \times 10^{-4} \, T$ with the magnetic moment vector making an angle $30^{\circ}$ with the field. Calculate the work done in increasing the angle from $30^{\circ}$ to $45^{\circ}$.