$A$ magnetic dipole of magnetic moment $ 6 \times 10^{-2} \text{ A m}^2 $ and moment of inertia $ 12 \times 10^{-6} \text{ kg m}^2 $ performs oscillations in a magnetic field of $ 2 \times 10^{-2} \text{ T} $. The time taken by the dipole to complete $ 20 $ oscillations is (assume $ \pi \simeq 3 $). (in $\text{ s}$)

Two identical short bar magnets,each having a magnetic moment of $10 \, Am^2$,are arranged such that their axial lines are perpendicular to each other and their centres lie along the same straight line in a horizontal plane. If the distance between their centres is $0.2 \, m$,find the resultant magnetic induction at a point midway between them. (Given: $\mu_0 = 4\pi \times 10^{-7} \, Hm^{-1}$)

The intensity of the magnetic field is $H$ and the magnetic moment of the magnet is $M$. The maximum potential energy is $..... MH$.

If a bar magnet of moment $10^{-4} Am^2$ is kept in a uniform magnetic field of $12 \times 10^{-3} T$ such that it makes an angle of $30^{\circ}$ with the direction of the magnetic field,then the torque acting on the magnet is:

$A$ magnet of magnetic moment $M$ is lying in a uniform magnetic field $B$. $W_1$ is the work done in turning it from $0^o$ to $60^o$ and $W_2$ is the work done in turning it from $30^o$ to $90^o$. Then:

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