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Chapter 20: Problem 10

There is associated with each atom in paramagnetic and ferromagnetic materials a net magnetic moment. Explain why ferromagnetic materials can be permanently magnetized whereas paramagnetic ones cannot.

Short Answer

Expert verified
Answer: Ferromagnetic materials can be permanently magnetized due to the aligned magnetic moments within magnetic domains and their ability to lock in place under a strong external magnetic field. In contrast, paramagnetic materials cannot be permanently magnetized because their magnetic moments become randomly oriented once the external magnetic field is removed, due to thermal energy.

Step by step solution

01

Understanding ferromagnetic and paramagnetic materials

Ferromagnetic materials, like iron, cobalt, and nickel, are strongly attracted to magnetic fields due to the presence of unpaired electrons. In these materials, the electron spins align parallel to each other, creating a strong magnetic field. In contrast, paramagnetic materials, such as aluminum and copper, are weakly attracted to magnetic fields. These materials also have unpaired electrons, but their spins align randomly, resulting in a weaker magnetic field.
02

Magnetic domains in ferromagnetic materials

Ferromagnetic materials consist of magnetic domains, which are regions where the magnetic moments of atoms are aligned parallel to each other. In the absence of an external magnetic field, these domains are randomly oriented, and the net magnetization is zero. However, when a ferromagnetic material is placed in an external magnetic field, these domains align with the external field, and the material becomes magnetized.
03

Behavior of paramagnetic materials under a magnetic field

When a paramagnetic material is placed in an external magnetic field, its magnetic moments temporarily align along the direction of the field, similar to what happens in ferromagnetic materials. However, once the external field is removed, thermal energy causes the magnetic moments to become randomly oriented again, resulting in the loss of the induced magnetization. Hence, paramagnetic materials cannot retain magnetization in the absence of a magnetic field.
04

Permanent magnetization of ferromagnetic materials

In ferromagnetic materials, if the external magnetic field is strong enough, the magnetic domains become reoriented and locked in place, even after the external field is removed. This process leads to a permanent magnetization of the material, as the new alignment of magnetic domains persists. In summary, ferromagnetic materials can be permanently magnetized due to the aligned magnetic moments within magnetic domains and their ability to lock in place under a strong external magnetic field. On the other hand, paramagnetic materials cannot be permanently magnetized because their magnetic moments become randomly oriented once the external magnetic field is removed, due to thermal energy.

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Most popular questions from this chapter

Confirm that there are \(2.2\) Bohr magnetons associated with each iron atom, given that the saturation magnetization is \(1.70 \times 10^{6} \mathrm{~A} / \mathrm{m}\), that iron has a BCC crystal structure, and that the unit cell edge length is \(0.2866 \mathrm{~nm}\).

The formula for yttrium iron garnet \(\left(\mathrm{Y}_{3} \mathrm{Fe}_{5} \mathrm{O}_{12}\right)\) may be written in the form \(\mathrm{Y}_{3}^{c} \mathrm{Fe}_{2}^{a} \mathrm{Fe}_{3}^{d} \mathrm{O}_{12}\), where the superscripts \(a, c\), and \(d\) represent different sites on which the \(\mathrm{Y}^{3+}\) and \(\mathrm{Fe}^{3+}\) ions are located. The spin magnetic moments for the \(\mathrm{Y}^{3+}\) and \(\mathrm{Fe}^{3+}\) ions positioned in the \(a\) and \(c\) sites are oriented parallel to one another and antiparallel to the \(\mathrm{Fe}^{3+}\) ions in \(d\) sites. Compute the number of Bohr magnetons associated with each \(\mathrm{Y}^{3+}\) ion, given the following information: (1) each unit cell consists of eight formula \(\left(\mathrm{Y}_{3} \mathrm{Fe}_{5} \mathrm{O}_{12}\right)\) units; (2) the unit cell is cubic with an edge length of \(1.2376 \mathrm{~nm} ;\) (3) the saturation magnetization for this material is. \(1.0 \times 10^{4} \mathrm{~A} / \mathrm{m}\); and (4) there are five Bohr magnetons associated with each \(\mathrm{Fe}^{3+}\) ion

An iron bar magnet having a coercivity of \(4000 \mathrm{~A} / \mathrm{m}\) is to be demagnetized. If the bar is inserted within a cylindrical wire coil \(0.15\) \(\mathrm{m}\) long and having 100 turns, what electric current is required to generate the necessary magnetic field?

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

A coil of wire \(0.1 \mathrm{~m}\) long and having 15 turns carries a current of \(1.0 \mathrm{~A}\). (a) Compute the flux density if the coil is within a vacuum. (b) A bar of an iron-silicon alloy, the \(B-H\) behavior for which is shown in Figure \(20.29\), is positioned within the coil. What is the flux density within this bar? (c) Suppose that a bar of molybdenum is now situated within the coil. What current must be used to produce the same \(B\) field in the Mo as was produced in the iron-
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