$A$ short bar magnet has a length $2l$ and a magnetic moment $10 \text{ Am}^2$. Find the magnetic field at a distance of $z = 0.1 \text{ m}$ from its centre on the axial line. Here,$l$ is negligible as compared to $z$.

Two identical thin bar magnets are placed mutually at right angles such that the north pole of one touches the south pole of the other. The length of each bar magnet is $\ell$. The magnitude of the resultant magnetic moment of the system is $[m = \text{pole strength of the magnet}]$.

If $L$ is the length of a bar magnet,then the separation between the two poles is nearly

$A$ 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.

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 magnetic field lines due to a bar magnet are correctly shown in