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.....$A{m^2}$

Write Gauss’s law for magnetism.

Following figures show the arrangement of bar magnets in different configurations. Each magnet has magnetic dipole moment $\vec m$ . Which configuration has highest net magnetic dipole moment

Two identical bar magnets are held perpendicular to each other with a certain separation, as shown below. The area around the magnets is divided into four zones. Given that there is a neutral point it is located in

The magnetic field at a point $x $ on the axis of a small bar magnet is equal to the field at a point $  y  $ on the equator of the same magnet. The ratio of the distances of $x$ and $y$  from the centre of the magnet is

Two magnets, each of magnetic moment $‘M’ $ are placed so as to form a cross at right angles to each other. The magnetic moment of the system will be