Question

Iron atom has magnetic moment 2.2 miuB.Calculate it's saturation magnetisation?

Answer 1

Answer on Question #67388, Physics / Solid State Physics

Iron atom has magnetic moment 2.2 miuB. Calculate it's saturation magnetisation?

Find: $M_s - ?$

Given:

$p = 2.2 \times 9.27 \times 10^{-24} \, \mathrm{J} \times \mathrm{T}^{-1}$

Solution:

Saturation magnetisation $M_s$ is a magnetic moment $p$ per unit volume $V$.

We find the magnetic moment per cubic cell (which contains 2 Fe atoms in bcc Fe metal) and divide this moment by the cell volume to obtain the magnetization in units of $\mathrm{emu/cm^3}$.

$$M_s = \frac{2 \times 2.2 \times 9.27 \times 10^{-21}}{(2.87 \times 10^{-24})^3} = 1725 \, \mathrm{emu/cm^3}$$

Answer:

$M_s = 1725 \, \mathrm{emu/cm^3}$

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