The average dipole moment of $Fe$ atoms is $1.8 × 10^{-23}\ A-m^2$ . The magnetic moment of an iron rod of length $10\ cm$ and diameter $1\,cm$ is........$A-m^2$ : (density and at. wt. of $Fe$ are $7.87\ g/cm^3$ and $55.87$ )
Medium
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Volume of domain
$=\pi\left(\frac{1}{100}\right)^{2} \times \frac{10}{100}=10 \pi$
mass $=7.87 \times 10 \pi$
$ = 78.7\,\pi \,{\text{gm}}$
${\text{N}} = \frac{{78.7 \times 2 \times 6.023 \times {{10}^{23}}}}{{7 \times (55.87)}}$
Magnetic moment $=\mathrm{N} \times 1.8 \times 10^{-23}$
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