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Chapter 32: Problem 51

The saturation magnetization \(M_{\max }\) of the ferromagnetic metal nickel is \(4.70 \times 10^{5} \mathrm{~A} / \mathrm{m} .\) Calculate the magnetic dipole moment of a single nickel atom. (The density of nickel is \(8.90 \mathrm{~g} / \mathrm{cm}^{3}\), and its molar mass is \(58.71 \mathrm{~g} / \mathrm{mol} .)\)

Short Answer

Expert verified
The magnetic dipole moment per nickel atom is approximately \(5.1512 \times 10^{-24} \text{ A} \cdot \text{m}^2\).

Step by step solution

01

Calculate the Volume of One Mole of Nickel

The molar mass of nickel is given as 58.71 g/mol. The density of nickel is 8.90 g/cm鲁. To find the volume of one mole of nickel, use the formula: \( \text{Volume} = \frac{\text{Molar Mass}}{\text{Density}} \). Substituting the values, we get:\[\text{Volume of 1 mole} = \frac{58.71 \text{ g/mol}}{8.90 \text{ g/cm}^3} = 6.60 \text{ cm}^3 / \text{mol}\]
02

Convert Volume to m鲁

Next, we need to convert the volume from cm鲁 to m鲁 since the magnetic saturation is in \( \text{A/m} \).\[1 \text{ cm}^3 = 1 \times 10^{-6} \text{ m}^3 \]Thus, the volume of one mole of nickel in \( \text{m}^3 \) is:\[6.60 \text{ cm}^3 / \text{mol} \times 1 \times 10^{-6} \text{ m}^3/\text{cm}^3 = 6.60 \times 10^{-6} \text{ m}^3 / \text{mol}\]
03

Calculate the Number of Atoms in One Mole

One mole of any substance contains Avogadro's number of atoms, which is approximately \( 6.022 \times 10^{23} \) atoms/mol. So, the number of nickel atoms in one mole is:\[6.022 \times 10^{23} \text{ atoms/mol}\]
04

Determine the Volume per Atom

Now we calculate the volume of nickel occupied by one atom by dividing the volume of one mole by the number of atoms in a mole:\[\text{Volume per atom} = \frac{6.60 \times 10^{-6} \text{ m}^3 / \text{mol}}{6.022 \times 10^{23} \text{ atoms/mol}}\]\[= 1.096 \times 10^{-29} \text{ m}^3/\text{atom}\]
05

Calculate Magnetic Dipole Moment per Atom

The saturation magnetization \( M_{\max} \) is related to the magnetic dipole moment \( \mu \) as: \[M_{\max} = \frac{N \cdot \mu}{V} \]where \( N \) is the number of atoms, and \( V \) is the total volume. We rearrange the equation to solve for \( \mu \):\[\mu = M_{\max} \cdot \text{Volume per atom}\]Substitute with \( M_{\max} = 4.70 \times 10^{5} \text{ A/m} \):\[\mu = 4.70 \times 10^{5} \text{ A/m} \times 1.096 \times 10^{-29} \text{ m}^3\]\[= 5.1512 \times 10^{-24} \text{ A} \cdot \text{m}^2\]

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Ferromagnetism
Ferromagnetism is a fascinating physical phenomenon observed in certain materials like nickel, iron, and cobalt. These materials exhibit a strong form of magnetism. This occurs when the magnetic dipoles within their atomic structure align in the same direction. This alignment happens even without an external magnetic field.
Ferromagnetic materials have unique properties. One is the ability to retain a magnetic state indefinitely after being magnetized. Another key property is their characteristic spontaneous magnetization, which is the net magnetic moment occurring in the absence of an externally applied magnetic field.
  • **Domains**: Ferromagnetic materials are divided into small regions known as domains. In each domain, the magnetic moments are aligned, maximizing their magnetic effect.
  • **Hysteresis**: This term describes the lag between an applied magnetic field and the material's magnetization. It's an important concept to understand how ferromagnetic materials work in devices like transformers and hard drives.
In summary, the remarkable strength of ferromagnets and their ability to preserve magnetic domains makes them valuable in various technologies.
Saturation Magnetization
Saturation magnetization is a critical concept in understanding the behavior of ferromagnetic materials. It represents the maximum magnetization that a ferromagnetic material can achieve. When all the magnetic dipoles become maximally aligned, the material reaches its saturation point.
Saturation occurs when an external magnetic field is strong enough to align all the magnetic moments in a material. Beyond this point, increasing the field does not result in additional magnetization. The material's response is then dictated by its magnetic saturation limit.
  • **Unit of Measurement**: The saturation magnetization is measured in amperes per meter (A/m).
  • **Importance**: It provides insight into the magnetic strength and capacity of materials, crucial for the design of magnetic devices. For instance, in the case of nickel, its saturation magnetization is measured as \(4.70 \times 10^{5} \mathrm{~A/m}\).
Saturation magnetization is an intrinsic property of a material and remains constant for a given material at a given temperature. Understanding this concept is pivotal for designing and optimizing magnetic systems.
Nickel
Nickel, a commonly known metal, exhibits interesting magnetic properties due to its ferromagnetic nature. It belongs to the transition metals on the periodic table and has the chemical symbol Ni.
Nickel's magnetic characteristics make it highly valuable in various technological applications. One remarkable quality of nickel is its saturation magnetization, which is \(4.70 \times 10^{5} \mathrm{~A/m}\), a reflection of its ability to magnetize strongly.
  • **Applications**: Because of its high magnetic permeability and resistance to corrosion, nickel is widely used in alloys and battery components.
  • **Abundance**: Nickel is the fifth most abundant element on Earth, which makes it accessible for industrial applications.
  • **Versatile Metal**: Apart from its magnetic qualities, nickel is also valued for its durability, making it suitable for coins and protective coatings.
Understanding nickel's role as a ferromagnetic material can help harness its potential in fields like electronics and metallurgy.

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Most popular questions from this chapter

A parallel-plate capacitor with circular plates of radius \(R\) is being charged. Show that the magnitude of the current density of the displacement current is \(J_{d}=\varepsilon_{0}(d E / d t)\) for \(r \leq R\).

The magnitude of the dipole moment associated with an atom of iron in an iron bar is \(2.1 \times 10^{-23} \mathrm{~J} / \mathrm{T}\). Assume that all the atoms in the bar, which is \(5.0 \mathrm{~cm}\) long and has a cross-sectional area of \(1.0 \mathrm{~cm}^{2},\) have their dipole moments aligned. (a) What is the dipole moment of the bar? (b) What torque must be exerted to hold this magnet perpendicular to an external field of magnitude \(1.5 \mathrm{~T} ?\) (The density of iron is \(\left.7.9 \mathrm{~g} / \mathrm{cm}^{3} .\right)\)

If an electron in an atom has orbital angular momentum with \(m_{e}\) values limited by 卤3 , how many values of (a) \(L_{\text {orb }, z}\) and (b) \(\mu_{\text {orb }, z}\) can the electron have? In terms of \(h, m,\) and \(e,\) what is the greatest allowed magnitude for (c) \(L_{\text {orb }, z}\) and \(\left(\right.\) d) \(\mu_{\text {orb }, z} ?\) (e) What is the greatest allowed magnitude for the \(z\) component of the electron's net angular momentum (orbital plus spin)? (f) How many values (signs included) are allowed for the \(z\) component of its net angular momentum?

Assume that an electron of mass \(m\) and charge magnitude \(e\) moves in a circular orbit of radius \(r\) about a nucleus. A uniform magnetic field \(\vec{B}\) is then established perpendicular to the plane of the orbit. Assuming also that the radius of the orbit does not change and that the change in the speed of the electron due to field \(\vec{B}\) is small, find an expression for the change in the orbital magnetic dipole moment of the electron due to the field.

A Rowland ring is formed of ferromagnetic material. It is circular in cross section, with an inner radius of \(5.0 \mathrm{~cm}\) and an outer radius of \(6.0 \mathrm{~cm},\) and is wound with 400 turns of wire. (a) \(\underline{W h a t}\) current must be set up in the windings to attain a toroidal field of magnitude \(B_{0}=0.20 \mathrm{mT} ?\) (b) A secondary coil wound around the toroid has 50 turns and resistance \(8.0 \Omega .\) If, for this value of \(B_{0},\) we have \(B_{M}=800 B_{0}\), how much charge moves through the secondary coil when the current in the toroid windings is turned on?
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