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Chapter 11: Problem 4

The Magnetic Moment of a Rotating Charged Disk A compact disk with radius \(R\) and uniform surface charge density \(\sigma\) rotates with angular speed \(\omega\). Find the magnetic dipole moment \(\mathbf{m}\) when the axis of rotation is (a) the symmetry axis of the disk; (b) any diameter of the disk.

Short Answer

Expert verified
(a) \( m = \frac{\pi \sigma \omega R^4}{4} \); (b) \(0\).

Step by step solution

01

Define the Problem and Given Information

We need to find the magnetic dipole moment for a rotating charged disk under two scenarios: when the rotation is about the symmetry axis and when it's about a diameter. The disk has radius \(R\), uniform surface charge density \(\sigma\), and rotates with angular speed \(\omega\).
02

Magnetic Moment about the Symmetry Axis (Part a)

For a disk rotating about its symmetry axis, consider a small ring element at a distance \(r\) from the center. The charge of this ring element is \(dq = \sigma (2 \pi r \, dr)\). The current \(I\) is due to this charge moving in a circular path, given by \(I = \frac{dq}{T} = \sigma \omega r \, dr\). The magnetic moment \(dm\) for this ring is \(dm = I \cdot \pi r^2 = \pi \sigma \omega r^3 \, dr\). Integrate from 0 to \(R\) to find the total magnetic moment: \[ m = \int_0^R \pi \sigma \omega r^3 \, dr = \frac{\pi \sigma \omega R^4}{4}. \]
03

Magnetic Moment about Any Diameter (Part b)

When the disk rotates about any diameter, the system is symmetric, causing contributions to the magnetic moment from symmetric portions of the disk to cancel out. Therefore, the net magnetic moment about this axis is \(0\).
04

Consolidate Results

From step 2, the magnetic dipole moment when the disk rotates about the symmetry axis is \(m = \frac{\pi \sigma \omega R^4}{4}\). From step 3, the magnetic moment when rotating about any diameter is \(0\).

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Key Concepts

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

Rotating Charged Disk
A rotating charged disk is a fascinating concept. Imagine a disk that has a radius, filled with an even distribution of electric charges on its surface. This distribution of charges gives it a uniform surface charge density. When the disk rotates, it behaves similarly to an electric current loop. This is because the motion of the charge creates a circular path. Each small section of the disk, as it rotates, can be thought of as a tiny loop of charge. This combination of rotation and charge gives rise to what we call a magnetic dipole moment. This is a measure of the strength of a magnetic source formed by the rotating charges. The problem is particularly interesting when we consider rotating this disk about different axes. Depending on how the disk rotates 鈥� either about its symmetry axis or a diameter 鈥� the resulting magnetic dipole moment can vary significantly. In fact, for a simple disk with uniform charge distribution, one rotation direction will produce a magnetic moment, while rotating in another will result in no magnetic moment at all! This encompasses the essence of rotational symmetry and charge dynamics.
Surface Charge Density
Surface charge density, denoted typically by \(\sigma\), refers to how much electric charge exists per unit area on a surface. For a rotating charged disk, surface charge density is a vital parameter. It tells us how thickly or sparsely the electric charges are spread over the disk's surface. This is crucial because the degree of surface charge density directly influences the resultant magnetic dipole moment when the disk rotates.Imagine if you painted electrons on a disk 鈥� the more densely packed these electrons, the higher the surface charge density \(\sigma\). This directly impacts our calculations. When you calculate how much charge a small ring on the disk's surface has, you rely on its surface charge density and the area of the ring to determine \(dq\), the differential charge. Moreover, this density, combined with the disk's rotation, dictates how intense the magnetic effects will be. More charge density means stronger potential magnetic effects from the rotating disk!
Symmetry Axis
The symmetry axis is essentially an imaginary line around which the disk is perfectly balanced. For a rotating charged disk, this is typically the line that runs perpendicular to the center of the disk, passing through its middle. Think of it as the axis of a spinning top. When the disk rotates about this symmetry axis, every point at a given distance from this axis has an identical movement path. This creates a situation where the charges move in perfect circles, generating continuous loops of current. Due to its symmetrical nature, this setup generates a straightforward scenario mathematically. Each small section's contribution to the overall magnetic dipole moment can be summed up to find a simple and concise expression for the total magnetic moment. This is why integrating over the radius is so effective in calculating the disk's magnetic dipole moment when spun about this axis.
Angular Speed
Angular speed, denoted by \(\omega\), measures how quickly an object rotates around an axis. It's expressed in units like radians per second. In the case of a rotating charged disk, angular speed is crucial. It essentially dictates the "rotational velocity" of each point on the disk's surface.A higher angular speed means each point on the disk traverses around its circular path more quickly. This increased velocity of charge movement boosts the "current" effect across the disk. Hence, as the angular speed of the disk increases, so does the magnetic dipole moment. This relationship manifests in our equations. For instance, in finding the magnetic moment due to a tiny ring of charge on the disk, the current \(I\) depends directly on \(\omega\). As such, the expression for the magnetic dipole moment incorporates \(\omega\) to reflect the influence of angular speed on the system's magnetic characteristics. This is why understanding angular speed gives insight into the magnitude of magnetic effects for rotating charged systems.

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Most popular questions from this chapter

Magnetic Dipole and Quadrupole Moments for \(\psi(\mathbf{r}) \quad\) The text produced a spherical multipole expansion for the magnetic scalar potential \(\psi(\mathbf{r})\) based on the identity $$ -r \frac{\partial \psi}{\partial r}=I=\frac{\mu_{0}}{4 \pi} \int d^{3} r^{\prime} \frac{\mathbf{r}^{\prime} \cdot \nabla^{\prime} \times \mathbf{j}\left(\mathbf{r}^{\prime}\right)}{\left|\mathbf{r}-\mathbf{r}^{\prime}\right|} $$ A Cartesian expansion for the scalar potential can be developed from the same starting point. (a) Expand \(\left|\mathbf{r}-\mathbf{r}^{\prime}\right|^{-1}\) and confirm that $$ \psi(\mathbf{r})=\frac{\mu_{0}}{4 \pi}\left[-\mathbf{m} \cdot \nabla \frac{1}{r}+\frac{1}{3} m_{i j}^{(2)} \nabla_{i} \nabla_{j} \frac{1}{r}+\cdots\right] $$ \(a\) $$ \mathbf{m}=\frac{1}{2} \int d^{3} r(\mathbf{r} \cdot \nabla \times \mathbf{j}) \mathbf{r} $$ $$ m_{i j}^{(2)}=\frac{1}{2} \int d^{3} r(\mathbf{r} \cdot \nabla \times \mathbf{j}) r_{i} r_{j} $$ where (b) Show that the dipole moment can be written in the familiar form \(\mathbf{m}=\frac{1}{2} \int d^{3} r \mathbf{r} \times \mathbf{j}\). (c) Show that the magnetic quadrupole can be written as $$ m_{i j}^{(2)}=\frac{1}{2} \int d^{3} r(\mathbf{r} \times \mathbf{j}) \cdot \nabla\left(r_{i} r_{j}\right) $$ (d) Prove that \(m_{i j}^{(2)}\) in part (c) is traceless. (e) Rewrite the formula in part (c) as \(\mathbf{m}^{(2)}=\frac{1}{2} \int d^{3} r[(\mathbf{r} \times \mathbf{j}) \mathbf{r}+\mathbf{r}(\mathbf{r} \times \mathbf{j})]\). (f) Prove that the magnetic quadrupole moment \(\mathbf{M}^{(2)}=\int d^{3} r(\mathbf{r} \times \mathbf{j}) \mathbf{r}\) produces the same quadrupole scal: potential as \(\mathbf{m}^{(2)}\) in part (c). \(\mathbf{M}^{(2)}\) is the moment which appears in the Cartesian multipole expansion \(\mathrm{f}\) the vector potential derived in the text.

A Hidden Delta Function The text writes two expressions for the magnetic field of point magnetic dipole at the origin: $$ \begin{aligned} &\mathbf{B}(\mathbf{r})=\mu_{0} \mathbf{m} \delta(\mathbf{r})-\nabla \psi_{0}(\mathbf{r}) \\ &\mathbf{B}(\mathbf{r})=\frac{\mu_{0}}{4 \pi} \frac{3 \hat{\mathbf{r}}(\hat{\mathbf{r}} \cdot \mathbf{m})-\mathbf{m}}{r^{3}}+\frac{2}{3} \mu_{0} \mathbf{m} \delta(\mathbf{r}) \end{aligned} $$ \(\psi_{0}(\mathbf{r})\) is the magnetic scalar potential of a point dipole. Prove that the delta function content of these two formulae are the same, at least when used as part of an integration over a volume integral.

The Field inside a Semi-Infinite Solenoid A semi-infinite solenoid (concentric with the negative \(z\)-axis) has cross sectional area \(A=\pi R^{2}, n\) turns per unit length of a wire with current \(I\), and magnetic moment per unit length \(\mathrm{m}=n I A \hat{\mathbf{z}}\). When \(A \rightarrow 0\) and \(N \rightarrow \infty\) with \(\mathrm{m}\) fixed, Application \(11.1\) showed that the magnetic field outside this solenoid (i.e., excluding the negative \(z\)-axis) is $$ \mathbf{B}_{o s}=\frac{\mu_{0} m}{4 \pi} \frac{\mathbf{r}}{r^{2}} . $$ Use the field inside an infinite solenoid to approximate the field \(\mathbf{B}_{\text {in }}\) inside a semi-infinite, finite-area solenoid. Show that when \(A \rightarrow 0, \mathbf{B}_{\text {in }}\) is confined to the negative \(z\)-axis and \(\nabla \cdot \mathbf{B}_{\text {in }}=-\mu_{0} m \delta(\mathbf{r})\). This guarantees that \(\nabla \cdot\left(\mathbf{B}_{\text {in }}+\mathbf{B}_{o t}\right)=0\) as it must. Hint: Show that \(\lim _{A \rightarrow 0} \Theta(R-\rho) / A=\delta(x) \delta(y)\).

18 Octupoles from Dipoles Let \(\mathbf{m}_{1}, \mathbf{m}_{2}\), and \(\mathbf{m}_{3}\) be three point dipoles. (a) Find the constraints that must be imposed on the \(\mathbf{m}_{\alpha}\) and their positions \(\mathbf{r}_{\alpha}\) if the asymptotic magnetic field is octupolar. (b) Show that one class of solutions places all the dipoles (with suitable magnitudes) on a line (separated by suitable distances) with their moments pointing parallel or anti-parallel to one another. Sketch an example. (c) Show that another class of solutions has all the dipoles at the corners of a triangle and, if \(s_{\alpha}\) is the length of the side opposite \(\mathbf{m}_{\alpha}, m_{1}: m_{2}: m_{2}=s_{1}: s_{2}: s_{3}\). Hint: The quadratic form \(x_{i} A_{i j} x_{j}=0\) if \(A_{i j}=-A_{j i}\).

Magnetic Dipole Moment Practice A current distribution produces the vector potential $$ \mathbf{A}(r, \theta, \phi)=\hat{\phi} \frac{\mu_{0}}{4 \pi} \frac{A_{0} \sin \theta}{r} \exp (-\lambda r) . $$ What is the magnetic moment associated with this current distribution?
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