A small bar magnet has a magnetic moment $1.2 \,A-m^2$. The magnetic field at a distance $ 0.1\, m $ on its axis will be : ($\mu_0 = 4\pi \times 10^{-7} \,T-m/A$)

A bar magnet of magnetic moment $M$ is cut into two parts of equal length. The magnetic moment of each part will be ......... $M$

Each atom of an iron bar $(5\,cm \times 1\,cm \times 1\,cm)$ has a magnetic moment $1.8 \times {10^{ - 23}}\,A{m^2}.$ Knowing that the density of iron is $7.78 \times {10^3}\,k{g^{ - 3}}\,m,$ atomic weight is $56$ and Avogadro's number is $6.02 \times {10^{23}}$ the magnetic moment of bar in the state of magnetic saturation will be.....$A{m^2}$

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?

Show the magnetic field of bar magnet in a panoramic way.

Two identical short bar magnets, each having magnetic moment of $10\, Am^2$, are arranged such that their axial lines are perpendicular to each other and their centres be along the same straight line in a horizontal plane. If the distance between their centres is $0.2 \,m$ , the resultant magnetic induction at a point midway between them is$({\mu _0} = 4\pi \times {10^{ - 7}}\,H{m^{ - 1}})$