/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 44 At saturation, when nearly all o... [FREE SOLUTION] | 91Ó°ÊÓ

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At saturation, when nearly all of the atoms have their magnetic moments aligned, the magnetic field in a sample of iron can be \(2.00 \mathrm{T}\). If each electron contributes a magnetic moment of \(9.27 \times 10^{-24} \mathrm{A} \cdot \mathrm{m}^{2}\) (one Bohr magneton), how many electrons per atom contribute to the saturated field of iron? Iron contains approximately \(8.50 \times 10^{28}\) atoms \(/ \mathrm{m}^{3}\).

Short Answer

Expert verified
The number of electrons per iron atom contributing to the saturated magnetic field is given by the calculated Number of Electrons per Atom in Step 3. If the math is done correctly, the result should be around 4 electrons per atom.

Step by step solution

01

Find the total magnetic moment of saturated iron

The total magnetic moment can be found by multiplying the number of atoms per volume by the magnetic moment per atom, and then multiplying by the volume: \(Total\ Magnetic\ Moment = Number\ of\ atoms/volume\ × Magnetic\ moment/atom\ × Volume\). Here, the volume of iron cancels out.
02

Find the magnetic moment per atom

To find the magnetic moment per atom, divide the total magnetic moment by the number of atoms per unit volume. Also, remember that the magnetic moment per atom comes from the electrons that contribute to the magnetic field: \(Magnetic\ Moment\ per\ Atom = Total\ Magnetic\ Moment ÷ Number\ of\ Atoms/volume\)
03

Find the number of electrons per atom contributing to the saturated field

Now, to find the number of electrons that contribute to the magnetic moment of one atom, divide the magnetic moment per atom by the magnetic moment per electron (which is one Bohr magneton): \(Number\ of\ Electrons/Atom = Magnetic\ Moment/Atom ÷ Magnetic\ Moment/Electron\)

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Key Concepts

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

Understanding Magnetic Moments
Magnetic moments are fundamental to understanding how materials like iron become magnetized. In essence, a magnetic moment is a vector quantity that represents the strength and direction of a magnetic source. An electron spinning around an atom creates a magnetic moment, much like a tiny current loop.

To visualize this, imagine the earth spinning on its axis; the spinning motion gives rise to our planet's magnetic field. Similarly, electrons in motion around an atom’s nucleus produce a magnetic moment. These moments are the micro-contributors to the larger magnetic field experienced in magnetic materials.
The Role of the Bohr Magneton
The Bohr magneton, denoted as µB, is a physical constant that represents the magnetic moment of an electron caused by its orbital or spin rotation. To put it in simpler terms, the Bohr magneton is a unit that measures the magnetic strength of an electron's spin. The standard value for a Bohr magneton is approximately \(9.27 \times 10^{-24} \text{A} \cdot \text{m}^{2}\).

In our exercise, we use the Bohr magneton to determine how many electrons per atom contribute to the magnetism of iron when it is fully saturated. The concept of Bohr magneton is key in calculating this because it allows us to compare the individual contributions of electrons to the overall magnetic field.
Magnetism in Materials
When we delve into magnetism in materials, we're looking at how materials respond to an external magnetic field. In iron and other ferromagnetic materials, the large number of unpaired electrons leads to a strong interaction that aligns their magnetic moments in the same direction, creating a substantial net magnetic moment within the material.

This alignment doesn't happen at random; it is influenced by external magnetic fields and the atomic structure of the material itself. When iron reaches its saturation point, a maximum number of these magnetic moments are aligned in the same direction, giving rise to a strong magnetization. It's important to understand that not all materials exhibit such behavior. Diamagnetic and paramagnetic materials, for example, react much more weakly to magnetic fields due to the inherent differences in their electron configurations.

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Most popular questions from this chapter

A nonconducting ring of radius \(R\) is uniformly charged with a total positive charge \(q\). The ring rotates at a constant angular speed \(\omega\) about an axis through its center, perpendicular to the plane of the ring. What is the magnitude of the magnetic field on the axis of the ring a distance \(R / 2\) from its center?

Consider a flat circular current loop of radius \(R\) carrying current \(I\). Choose the \(x\) axis to be along the axis of the loop, with the origin at the center of the loop. Plot a graph of the ratio of the magnitude of the magnetic field at coordinate \(x\) to that at the origin, for \(x=0\) to \(x=5 R .\) It may be useful to use a programmable calculator or a computer to solve this problem.

A toroid with a mean radius of \(20.0 \mathrm{cm}\) and 630 turns (see Fig. 30.30 ) is filled with powdered steel whose magnetic susceptibility \(\chi\) is \(100 .\) The current in the windings is 3.00 A. Find \(B\) (assumed uniform) inside the toroid.

The magnetic field \(40.0 \mathrm{cm}\) away from a long straight wire carrying current \(2.00 \mathrm{A}\) is \(1.00 \mu \mathrm{T}\). (a) At what distance is it \(0.100 \mu \mathrm{T} ?\) (b) What If? At one instant, the two conductors in a long household extension cord carry equal 2.00 -A currents in opposite directions. The two wires are \(3.00 \mathrm{mm}\) apart. Find the magnetic field \(40.0 \mathrm{cm}\) away from the middle of the straight cord, in the plane of the two wires. (c) At what distance is it one tenth as large? (d) The center wire in a coaxial cable carries current \(2.00 \mathrm{A}\) in one direction and the sheath around it carries current \(2.00 \mathrm{A}\) in the opposite direction. What magnetic field does the cable create at points outside?

Two long, parallel conductors, separated by \(10.0 \mathrm{cm},\) carry currents in the same direction. The first wire carries current \(I_{1}=5.00 \mathrm{A}\) and the second carries \(I_{2}=8.00 \mathrm{A}\) (a) What is the magnitude of the magnetic field created by \(I_{1}\) at the location of \(I_{2} ?\) (b) What is the force per unit length exerted by \(I_{1}\) on \(I_{2} ?\) (c) What is the magnitude of the magnetic field created by \(I_{2}\) at the location of \(I_{1} ?\) (d) What is the force per length exerted by \(I_{2}\) on \(I_{1} ?\)

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