$A$ 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?

Two identical bar magnets of magnetic moment $M$ each, are placed along $X$ and $Y$-axes, respectively at a distance $d$ from the origin (as shown in the figure). The origin lies on the perpendicular bisector of the magnet placed on the $X$-axis and on the magnetic axis of the magnet placed on the $Y$-axis. If the magnitude of the total magnetic field at the origin is $B = \alpha \left[ \frac{\mu_0}{4 \pi} \frac{M}{d^3} \right]$, then the value of the constant $\alpha$ will be (given $d >> l$, where $l$ is the length of the bar magnets and the direction of $N$ to $S$ in the magnets is opposite with respect to each other).

The ratio of magnetic intensities for an axial point and a point on the broad side-on (equatorial) position at an equal distance $d$ from the centre of a short bar magnet is:

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:

Two short magnetic dipoles $m_{1}$ and $m_{2}$, each having a magnetic moment of $1\, \text{Am}^{2}$, are placed at points $O$ and $P$ respectively. The distance between $O$ and $P$ is $1\, \text{m}$. The torque experienced by the magnetic dipole $m_{2}$ due to the presence of $m_{1}$ is ...... $\times 10^{-7}\, \text{Nm}$.

$A$ bar magnet with a magnetic moment $5.0\,Am^2$ is placed in a parallel position relative to a magnetic field of $0.4\,T$. The amount of work done in turning the magnet from a parallel to an antiparallel position relative to the field direction is $.........\,J$.