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

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

The incorrect statement regarding the lines of force of the magnetic field $B$ is:

$A$ bar magnet having pole strength $q_m$ and magnetic moment $m$ is divided into two equal parts along its length. The new pole strength is . . . . . . and the magnetic moment is . . . . . . respectively.

Inside a bar magnet,the magnetic lines of force

$A$ steel wire has a magnetic moment $M$. What will be its magnetic moment if it is bent into a semi-circular arc?