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Chapter 32: Problem 38

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.

Short Answer

Expert verified
The change in magnetic moment is \( \Delta \mu = \frac{e^2 r^2 B}{2m} \).

Step by step solution

01

Identify Initial Magnetic Moment

The magnetic moment \( \mu \) of an electron in a circular orbit is given by \( \mu = \frac{e v r}{2} \), where \( e \) is the electron charge, \( v \) is the velocity, and \( r \) is the radius of the orbit.
02

Relate Velocity Changes to Orbital Frequency

The electron's velocity is related to the angular frequency \( \omega \) by the equation \( v = \omega r \). When a magnetic field \( \vec{B} \) is introduced, the change in angular frequency \( \Delta \omega \) causes a small change in velocity \( \Delta v = r \Delta \omega \).
03

Determine Change in Magnetic Moment

The change in the magnetic moment \( \Delta \mu \) can be calculated from the change in angular frequency. Since \( \Delta \mu = \frac{e \Delta v r}{2} \), substituting \( \Delta v = r \Delta \omega \) gives \( \Delta \mu = \frac{e r^2 \Delta \omega}{2} \).
04

Calculate Change in Angular Frequency

In the presence of the magnetic field, the electron experiences a Lorentz force that changes its dynamics such that \( e B = m \Delta \omega \). Solving for \( \Delta \omega \), we find \( \Delta \omega = \frac{e B}{m} \).
05

Express Change in Magnetic Moment in Terms of B

Using \( \Delta \omega = \frac{e B}{m} \) in the expression for \( \Delta \mu \) gives \( \Delta \mu = \frac{e r^2}{2} \left( \frac{e B}{m} \right) = \frac{e^2 r^2 B}{2m} \).

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Key Concepts

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

Magnetic Field Effects
When a magnetic field interacts with moving charges, such as electrons, it can significantly affect their behavior. In this exercise, an electron orbits a nucleus in a circular path within the influence of a magnetic field perpendicular to the orbit. This field exerts a Lorentz force on the electron, leading to a change in its motion.
  • The electron experiences a force directly proportional to the product of its charge, velocity, and the magnetic field strength.
  • This force is perpendicular to the direction of motion and causes the electron's path to curve, affecting its velocity and frequency, thus influencing the orbital magnetic dipole moment.
Understanding these effects helps in predicting changes to the magnetic properties of electrons in a magnetic field.
Electron Dynamics
Electron dynamics in a magnetic field play a crucial role in understanding the electron's movement and its associated properties. When electrons move in an orbit subjected to a magnetic field, dynamic changes occur which involve the alteration of its angular velocity.
  • Electrons deviate from their normal path due to the magnetic field acting as a centripetal force.
  • This deviation adjusts the angular velocity of the electron, thereby impacting its orbital frequency.
    • Angular velocity is how fast the electron moves around the nucleus.
  • The electron maintains its orbit radius, despite experiencing a slight change in speed, due to the field.
These dynamics are pivotal for calculating the changes in the orbital magnetic dipole moment.
Angular Frequency Change
The introduction of a magnetic field causes a change in the angular frequency of the electron's movement. Angular frequency is the rate of rotation, and slight variations in this frequency can lead to significant modifications in the electron's magnetic moment.
  • The change in angular frequency is due to the Lorentz force, which modifies the kinetic energy of the electron, typically resulting in a small speed change.
  • The angular frequency change is given by \[\Delta \omega = \frac{e B}{m}\]relating it to electron charge, magnetic field strength, and electron mass, demonstrating why and how the frequency varies.
  • This frequency change is key to understanding how the orbital magnetic dipole moment of the electron adjusts when the magnetic field is applied.
    • The orbital magnetic dipole moment's adjustment is a direct consequence of the changes in how quickly the electron orbits.
These insights are essential when evaluating magnetic materials and their interactions in physical applications.

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Most popular questions from this chapter

A parallel-plate capacitor with circular plates of radius \(0.10\) m is being discharged. A circular loop of radius \(0.20 \mathrm{~m}\) is concentric with the capacitor and halfway between the plates. The displacement current through the loop is \(2.0 \mathrm{~A}\). At what rate is the electric field between the plates changing?

As a parallel-plate capacitor with circular plates \(20 \mathrm{~cm}\) in diameter is being charged, the current density of the displacement current in the region between the plates is uniform and has a magnitude of \(20 \mathrm{~A} / \mathrm{m}^{2}\). (a) Calculate the magnitude \(B\) of the magnetic field at a distance \(r=50 \mathrm{~mm}\) from the axis of symmetry of this region. (b) Calculate \(d E / d t\) in this region.

Earth has a magnetic dipole moment of \(8.0 \times 10^{22} \mathrm{~J} / \mathrm{T}\). (a) What current would have to be produced in a single turn of wire extending around Earth at its geomagnetic equator if we wished to set up such a dipole? Could such an arrangement be used to cancel out Earth's magnetism (b) at points in space well above Earth's surface or (c) on Earth's surface?

A magnetic compass has its needle, of mass \(0.050 \mathrm{~kg}\) and length \(4.0 \mathrm{~cm}\), aligned with the horizontal component of Earth's magnetic field at a place where that component has the value \(B_{h}=\) \(16 \mu \mathrm{T}\). After the compass is given a momentary gentle shake, the needle oscillates with angular frequency \(\omega=45 \mathrm{rad} / \mathrm{s}\). Assuming that the needle is a uniform thin rod mounted at its center, find the magnitude of its magnetic dipole moment.

A magnetic rod with length \(6.00 \mathrm{~cm}\), radius \(3.00 \mathrm{~mm}\), and (uniform) magnetization \(2.70 \times 10^{3} \mathrm{~A} / \mathrm{m}\) can turn about its center like a compass needle. It is placed in a uniform magnetic field \(\vec{B}\) of magnitude \(35.0 \mathrm{mT}\), such that the directions of its dipole moment and \(\vec{B}\) make an angle of \(68.0^{\circ}\). (a) What is the magnitude of the torque on the rod due to \(\vec{B} ?\) (b) What is the change in the orientation energy of the rod if the angle changes to \(34.0^{\circ} ?\)
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