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Chapter 20: Problem 25

Assume that the commercial iron \((99.95\) \(w t \% \mathrm{Fe}\) ) in Table \(20.5\) just reaches the point of saturation when inserted within the coil in Problem 20.1. Compute the saturation magnetization.

Short Answer

Expert verified
Question: Calculate the saturation magnetization of commercial iron with 99.95 wt% Fe when placed within a coil. Answer: The saturation magnetization of the commercial iron is 2.474 × 10^6 A/m.

Step by step solution

01

Calculate the number of Fe atoms in the given mass of commercial iron.

We are given 99.95 wt% Fe, and let's assume we have 1 kg of commercial iron, which means we have 0.9995 kg of Fe. Now we can calculate the number of Fe atoms using Avogadro's number. The atomic weight of Fe is 55.845 g/mol, so we need to find the number of moles first: Number of moles = mass / atomic weight Number of moles of Fe = 0.9995 kg / 55.845 g/mol = 17.905 mol Now we can find the number of Fe atoms using Avogadro's number (6.022 × 10^23 atoms/mol): Number of Fe atoms = 17.905 mol × 6.022 × 10^23 atoms/mol = 1.078 × 10^25 atoms
02

Calculate the volume of the given mass of commercial iron.

We need to find the volume of this 1 kg of commercial iron to calculate the magnetization. The density of commercial iron is 7.874 g/cm³. Volume = mass / density So the volume of the iron = mass of iron (in grams) / density of iron Volume = (0.9995 kg * 1000 g/kg) / 7.874 g/cm³ = 126.89 cm³
03

Calculate the number of unpaired electrons per Fe atom.

The electron configuration of Fe is [Ar] 3d6 4s2, which means there are 4 unpaired electrons in the 3d subshell.
04

Calculate the saturation magnetization.

To compute the saturation magnetization, we use the following formula: Saturation magnetization (Ms) = (number of unpaired electrons per atom × Bohr magneton × number of Fe atoms) / volume Bohr magneton (µB) = 9.274 × 10^(-24) Am² Ms = (4 × 9.274 × 10^(-24) Am² × 1.078 × 10^25 atoms) / 126.89 cm³ Converting the volume to m³, we have: Volume = 126.89 cm³ * 10^(-6) m³/cm³= 1.2689 × 10^(-4)m³ Now we can plug the volume into the formula and finally compute the saturation magnetization: Ms = (4 × 9.274 × 10^(-24) Am² × 1.078 × 10^25 atoms) / (1.2689 × 10^(-4)m³) Ms = 2.474 × 10^6 A/m Hence, the saturation magnetization of the commercial iron is 2.474 × 10^6 A/m.

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Most popular questions from this chapter

The following data are for a transformer steel: \begin{tabular}{cccc} \hline \multicolumn{3}{c}{\(\boldsymbol{B}\)} \\ \(\boldsymbol{H}(\mathrm{A} / \mathrm{m})\) & \((\) teslas \()\) & \(\boldsymbol{H}(\mathbf{A} / \mathrm{m})\) & \(\boldsymbol{B}\) (teslas) \\ \hline 0 & 0 & 200 & \(1.04\) \\ 10 & \(0.03\) & 400 & \(1.28\) \\ 20 & \(0.07\) & 600 & \(1.36\) \\ 50 & \(0.23\) & 800 & \(1.39\) \\ 100 & \(0.70\) & 1000 & \(1.41\) \\ 150 & \(0.92\) & & \\ \hline \end{tabular} (a) Construct a graph of \(B\) versus \(H\). (b) What are the values of the initial permeability and initial relative permeability? (c) What is the value of the maximum permeability? (d) At about what \(H\) field does this maximum permeability occur? (e) To what magnetic susceptibility does this maximum permeability correspond?

Cite the primary limitation of the new superconducting materials that have relatively high critical temperatures.

Estimate saturation values of \(H\) for singlecrystal iron in [100], [110], and [111] directions.

The magnetization within a bar of some metal alloy is \(3.2 \times 10^{5} \mathrm{~A} / \mathrm{m}\) at an \(H\) field of \(50 \mathrm{~A} / \mathrm{m}\). Compute the following: (a) the magnetic susceptibility, (b) the permeability, and (c) the magnetic flux density within this material. (d) What type(s) of magnetism would you suggest as being displayed by this material? Why?

A ferromagnetic material has a remanence of \(1.25\) teslas and a coercivity of \(50,000 \mathrm{~A} / \mathrm{m}\). Saturation is achieved at a magnetic field intensity of \(100,000 \mathrm{~A} / \mathrm{m}\), at which the flux density is \(1.50\) teslas. Using these data, sketch the entire hysteresis curve in the range \(H=-100,000 \mathrm{to}+100,000 \mathrm{~A} / \mathrm{m}\). Be sure to scale and label both coordinate axes.
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