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Chapter 32: Problem 42

An electron is placed in a magnetic field \(\vec{B}\) that is directed along a \(z\) axis. The energy difference between parallel and antiparallel alignments of the \(z\) -component of the electron's spin magnetic moment with \(\vec{B}\) is \(6.00 \times 10^{-25} \mathrm{~J}\). What is the magnitude of \(\vec{B}\) ?

Short Answer

Expert verified
The magnitude of the magnetic field is \(3.24 \times 10^{-2} \, T\).

Step by step solution

01

Understand the Problem

The problem involves an electron in a magnetic field \(\vec{B}\), and you're asked to find the magnitude of the magnetic field given the energy difference between parallel and antiparallel alignments of the electron's spin.
02

Recall the Formula

The energy difference \(\Delta E\) between the parallel and antiparallel alignments of the electron's spin magnetic moment in a magnetic field is given by \(\Delta E = 2 \mu_B B\), where \(\mu_B\) is the Bohr magneton and \(B\) is the magnitude of the magnetic field.
03

Substitute the Values

\(\mu_B\) (Bohr magneton) is a known constant: \(\mu_B = 9.27 \times 10^{-24} \, \mathrm{J/T}\).The given energy difference \(\Delta E\) is \(6.00 \times 10^{-25} \, \mathrm{J}\). Substitute these values into the formula: \(\Delta E = 2 \mu_B B\).
04

Solve for the Magnetic Field B

Rearrange the formula to solve for \(B\): \( B = \frac{\Delta E}{2 \mu_B}\).
05

Calculate the Result

Substitute the known values into the equation: \( B = \frac{6.00 \times 10^{-25} \, \mathrm{J}}{2 \times 9.27 \times 10^{-24} \, \mathrm{J/T}} = \frac{6.00 \times 10^{-25}}{1.854 \times 10^{-23}} \).Simplify the expression to find the magnitude of \(B\): \( B = 3.24 \times 10^{-2} \, T \).

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Key Concepts

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

Electron spin
An electron has a property known as spin, which is a fundamental characteristic of quantum particles. Spin can be thought of as a type of intrinsic angular momentum.
Electrons can spin in two opposite directions: either up (parallel to the magnetic field) or down (antiparallel to the magnetic field).
The spin of an electron creates a small magnetic moment, which means the electron behaves like a tiny magnet. This magnetic moment interacts with external magnetic fields, leading to different energy states.
In the presence of a magnetic field directed along the z-axis, the electron can align its spin either parallel or antiparallel to the field. These alignments correspond to different energy levels.
It's important to understand that the electron spin and the resulting magnetic moment are central to various physical phenomena, including the basics of magnetism and the operation of MRI machines.
Energy difference
The energy difference between the parallel and antiparallel alignments of an electron's spin in a magnetic field is critical for calculating the magnetic field magnitude.
When an electron's spin is parallel to the magnetic field, it is in a lower energy state compared to when it is antiparallel.
The energy associated with these states can be calculated using the formula \(\rightarrow E = 2 \mu_B B \), where \(\rightarrow 2 \mu_B B \) represents the energy difference (\(\rightarrow \Delta E \)) induced by the magnetic field B.
This equation focuses on the fact that the electron's spin magnetic moment interacts with the magnetic field, resulting in distinct energy levels.
Given that \(\rightarrow \Delta E \) for our example is \(\rightarrow 6.00 \times 10^{-25} \text{ J} \), you can substitute this value along with the Bohr magneton into the formula to find the magnitude of the magnetic field.
Bohr magneton
The Bohr magneton (\(\rightarrow \mu_B \)) is a physical constant that quantifies the magnetic moment of an electron due to its spin.
The value of the Bohr magneton is approximately \(\rightarrow 9.27 \times 10^{-24} \text{ J/T} \). This tiny value signifies that the magnetic moment is extremely small but still crucial for understanding magnetic interactions at the atomic level.
In our given formula, \(\rightarrow \Delta E = 2 \mu_B B \), \(\rightarrow \mu_B \) represents the proportionality constant that bridges the energy difference (due to magnetic interactions) and the magnetic field.
The factor of 2 accounts for the two possible orientations (parallel and antiparallel) of the electron's spin relative to the magnetic field.
Using the Bohr magneton, you can derive the magnitude of the magnetic field B by rearranging and solving the formula, as demonstrated in the provided solution. This reveals the field strength necessary to produce the observed energy difference.

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Most popular questions from this chapter

The magnetic field of the Earth can be approximated as the magnetic field of a dipole, with horizontal and vertical components, at a point a distance \(r\) from the Earth's center, given by $$ B_{h}=\frac{\mu_{0} \mu}{4 \pi r^{3}} \cos \lambda_{m}, \quad B_{v}=\frac{\mu_{0} \mu}{2 \pi r^{3}} \sin \lambda_{m}, $$ where \(\lambda_{m}\) is the magnetic latitude (this type of latitude is measured from the geomagnetic equator toward the north or south geomagnetic pole). Assume that the Earth's magnetic dipole moment is \(\mu=8.00 \times 10^{22} \mathrm{~A} \cdot \mathrm{m}^{2} .\) (a) Show that the magnitude of the Earth's field at latitude \(\lambda_{m}\) is given by $$ B=\frac{\mu_{0} \mu}{4 \pi r^{3}} \sqrt{1+3 \sin ^{2} \lambda_{m}} $$ (b) Show that the inclination \(\phi_{i}\) of the magnetic field is related to the magnetic latitude \(\lambda_{m}\) by $$ \tan \phi_{i}=2 \tan \lambda_{m} $$

Measurements in mines and boreholes indicate that the Earth's interior temperature increases with depth at the average rate of \(30 \mathrm{C}^{\circ} / \mathrm{km}\). Assuming a surface temperature of \(10^{\circ} \mathrm{C}\), at what depth does iron cease to be ferromagnetic? (The Curie temperature of iron varies very little with pressure.)

Two long parallel wires, both of radius \(a\) and whose centers are a distance \(d\) apart, carry equal currents in opposite directions. Show that, neglecting the flux within the wires, the inductance of a length \(l\) of such a pair of wires is given by $$ L=\frac{\mu_{0} l}{\pi} \ln \frac{d-a}{a} $$ (Hint: Calculate the flux through a rectangle of which the wires form two opposite sides.)

A transformer has 500 primary turns and 10 secondary turns. (a) If \(\Delta V_{p}\) is \(120 \mathrm{~V}\) (rms), what is \(\Delta V_{s}\) with an open circuit? (b) If the secondary now has a resistive load of \(15 \Omega\), what are the currents in the primary and secondary?

Two inductors \(L_{1}\) and \(L_{2}\) are connected in parallel and separated by a large distance. (a) Show that the equivalent inductance is given by $$ \frac{1}{L_{\mathrm{eq}}}=\frac{1}{L_{1}}+\frac{1}{L_{2}} $$ (Hint: Review the derivations for resistors in parallel and capacitors in parallel. Which is similar here?) (b) Why must their separation be large for this relationship to hold? (c) What is the generalization of (a) for \(N\) inductors in parallel?
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