The magnetic moment of a short magnet is $16 \ Am^2$. The magnetic induction at a point $20 \ cm$ away from its midpoint on $(i)$ axial point and $(ii)$ equatorial point respectively,will be:

$A$ donut-shaped permanent magnet (magnetization parallel to the axis) can slide frictionlessly on a vertical rod. Treat the magnets as dipoles with mass $m_d$ and dipole moment $M$. When we put two back-to-back magnets on the rod,the upper one will float. At what height $z$ does it float?

$A$ short bar magnet experiences a torque of magnitude $0.64 \ J$ when it is placed in a uniform magnetic field of $0.32 \ T$. The magnetic moment of the magnet is . . . . . .

$A$ short bar magnet of magnetic moment $5.25 \times 10^{-2} \;J\, T^{-1}$ is placed with its axis perpendicular to the earth's field direction. At what distance from the centre of the magnet,the resultant field is inclined at $45^{\circ}$ with the earth's field on
$(a)$ its normal bisector and
$(b)$ its axis.
Magnitude of the earth's field at the place is given to be $0.42 \;G$. Ignore the length of the magnet in comparison to the distances involved.

Due to a small magnet,the magnetic intensity at a distance $x$ in the end-on position (axial position) is $9 \ Gauss$. What will be the intensity at a distance $\frac{x}{2}$ in the broadside-on position (equatorial position)?

The magnitude of the axial field due to a short bar magnet at a distance of $50 \,cm$ from its mid-point is (The magnetic moment of the bar magnet is $0.4 \,Am^2$)