Compute (a) the saturation magnetization and (b) the saturation flux density for iron, which has a net magnetic moment per atom of 2.2 Bohr magnetons and a density of \(7.87 \mathrm{g} / \mathrm{cm}^{3}\).

Using the density and the molar mass of iron, we will compute the number of atoms per unit volume (n). The molar mass of iron is 55.845 g/mol, and Avogadro's number (N_A) is \(6.022 \times 10^{23} \text{atoms/mol}\). The density of iron is given as \(7.87 \text{g/cm}^3\). The formula to find the number of atoms per unit volume is: \(n = \frac{\rho N_A}{M}\), where \(\rho\) is the density, \(N_A\) is Avogadro's number, and \(M\) is the molar mass. \(n = \frac{7.87 \frac{\text{g}}{\text{cm}^3} \times 6.022 \times 10^{23} \frac{\text{atoms}}{\text{mol}}}{55.845 \frac{\text{g}}{\text{mol}}} \approx 8.489 \times 10^{22} \frac{\text{atoms}}{\text{cm}^3}\)

Now we can calculate the saturation magnetization (M_s) using the formula: \(M_s = n \times m\), where m is the net magnetic moment per atom given by \(2.2 \text{ Bohr magnetons}\). First, we need to convert the Bohr magnetons to Amperes per meter (\(\text{A/m}\)). One Bohr magneton is equal to \(\mu_B = 9.274 \times 10^{-24} \text{A m}^2\). \(m = 2.2 \text{ Bohr magnetons} \times \mu_B = 2.2 \times 9.274 \times 10^{-24} \text{A m}^2 = 20.402 \times 10^{-24} \text{A m}^2\) Now we can compute the saturation magnetization: \(M_s = n \times m = (8.489 \times 10^{22} \frac{\text{atoms}}{\text{cm}^3}) \times (20.402 \times 10^{-24} \text{A m}^2) \approx 1.732 \times 10^6 \text{ A/m}\) So the saturation magnetization for iron is approximately \(1.732 \times 10^6 \, \text{A/m}\).

Having found the saturation magnetization, we can compute the saturation flux density (B_s) using the formula: \(B_s = \mu_0 M_s\), where \(\mu_0\) is the permeability of free space which is equal to \(4\pi \times 10^{-7} \, \text{H/m}\). \(B_s = \mu_0 M_s = (4\pi \times 10^{-7} \, \text{H/m}) \times (1.732 \times 10^6 \, \text{A/m}) \approx 2.174 \times 10^{-3} \, \text{T}\) The saturation flux density for iron is approximately \(2.174 \times 10^{-3} \, \text{T}\). In conclusion, the saturation magnetization of iron is approximately \(1.732 \times 10^6 \, \text{A/m}\), and the saturation flux density is approximately \(2.174 \times 10^{-3} \, \text{T}\).