$A$ bar magnet of magnetic moment $1.5 \, A \cdot m^2$ lies aligned with the direction of a uniform magnetic field of $2 \, T$. What is the amount of work required to turn the magnet so as to align its magnetic moment normal to the field direction?

If a magnet of length $10 \, cm$ and pole strength $40 \, A-m$ is placed at an angle of $45^\circ$ in a uniform magnetic field of intensity $2 \times 10^{-4} \, T$,the torque acting on it is:

$A$ 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

$A$ short bar magnet of magnetic moment $10^4 \,J \,T^{-1}$ is free to rotate in a horizontal plane. The work done in rotating the magnet slowly from the direction parallel to a horizontal magnetic field of $4 \times 10^{-5} \,T$ to a direction $60^{\circ}$ to the direction of the field is (in $J$)

Write the equation of torque on a magnetic needle placed in a uniform magnetic field and derive the expression for its time period $T = 2\pi \sqrt{\frac{I}{mB}}$.

$A$ bar magnet of length $6\,cm$ has a magnetic moment of $4\,J\,T^{-1}$. Find the strength of magnetic field at a distance of $200\,cm$ from the centre of the magnet along its equatorial line.