$A$ 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 $...... \times 10^{-4} \, N \cdot m$.

$A$ bar magnet is hung by a thin cotton thread in a uniform horizontal magnetic field and is in equilibrium state. The energy required to rotate it by $60^{\circ}$ is $W$. Now,the torque required to keep the magnet in this new position is

If there is no torsion in the suspension thread, then the time period of a magnet executing $SHM$ is

The rate of change of torque $\tau$ with respect to deflection $\theta$ is maximum for a magnet suspended freely in a uniform magnetic field of induction $B$ when $\theta = ........ ^\circ$.

$A$ 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 equal to $4.5 \times 10^{-2} \; J$. What is the magnitude of magnetic moment (in $J \; T^{-1}$) of the magnet?

If a bar magnet of magnetic moment $M$ is freely suspended in a uniform magnetic field of strength $B$,the work done in rotating the magnet through an angle $\theta$ is