$A$ small bar magnet placed with its axis at $30^{\circ}$ with an external field of $0.06\, T$ experiences a torque of $0.018\, Nm$. The minimum work required to rotate it from its stable to unstable equilibrium position is

In which direction would the magnetic field on the axis at a distance $Z$ from the centre of the bar magnet be?

$A$ magnet of length $10 \text{ cm}$ and magnetic moment $1 \text{ Am}^2$ is placed along side $AB$ of an equilateral triangle $ABC$. If the length of the side $AB$ is $10 \text{ cm}$, find the magnetic induction at point $C$. (Given $\mu_0 = 4\pi \times 10^{-7} \text{ Hm}^{-1}$)

$A$ 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 $\text{ s}$)

$A$ bar magnet of magnetic moment $\overrightarrow{M}$ is placed in a magnetic field of induction $\overrightarrow{B}$. The torque exerted on it is

The work done in turning a magnet of magnetic moment $M$ by an angle of $90^{\circ}$ from the magnetic meridian is $n$ times the corresponding work done to turn it through an angle of $60^{\circ}$. The value of $n$ is: