The magnetic moment of a magnet is $2.5 \, J \, T^{-1}$. How much work in $J$ must be done to rotate it from its stable equilibrium position to an unstable equilibrium position in a magnetic field of $0.2 \, T$?

$A$ magnetic dipole is placed in a uniform magnetic field of intensity $B$,oriented along the direction of the field. If the magnetic dipole moment is $M$,then the maximum work an external agent can perform in rotating the dipole will be

The work done in turning a magnet of magnetic moment $M$ by an angle $\theta$ from its initial position parallel to the magnetic meridian is:

$A$ short bar magnet placed with its axis at $30^{\circ}$ to a uniform external magnetic field of $0.5 \ T$ experiences a torque of magnitude equal to $4.5 \times 10^{-2} \ J$. The magnitude of the magnetic moment is . . . . . . .

The magnitude of the axial field due to a bar magnet at a distance of $1 \ m$ is found to be $5 \times 10^{-8} \ T$. The magnetic moment of the bar magnet is $\left(\mu_0 = 4 \pi \times 10^{-7} \ T \ m/A\right)$. (in $A \ m^2$)

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