$A$ magnet,when placed perpendicular to a uniform magnetic field of strength $10^{-4} \ Wb/m^2$,experiences a maximum torque of $4 \times 10^{-5} \ N \cdot m$. What is its magnetic moment in $A \cdot m^2$?

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

$A$ bar magnet having centre $O$ has a length of $4 \ cm$. Point $P_1$ is in the broad side-on (equatorial) position and $P_2$ is in the end side-on (axial) position with $OP_1 = OP_2 = 10 \ m$. The ratio of magnetic intensities $H$ at $P_1$ and $P_2$ is:

Assertion: The poles of a magnet cannot be separated by breaking it into two pieces.
Reason: The magnetic moment will be reduced to half when a magnet is broken into two equal pieces.

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