Question

The magnetic moment $\overrightarrow{\boldsymbol{\mu}}$ of a spinning object with charge $Q$ and mass $M$ is proportional to its angular momentum $\vec{L}$ according to $\overrightarrow{\boldsymbol{\mu}}=g(Q / 2 M) \overrightarrow{\boldsymbol{L}} .$ The dimensionless coefficient $g$ is known as the $g$ -factor of the object. Consider a spherical shell with radius $R$ and uniform charge density $\sigma$ spinning with angular velocity $\omega$ around the $z$ -axis. (a) What is the differential area da of the ring at latitude $\theta,$ with width $R d \theta$, as shown in Fig. $\mathrm{P} 41.57 ?$ (b) The current $d l$ carried by the ring is its charge $\sigma$ da divided by the period of rotation. Determine $d I$ in terms of $R, \sigma, \omega,$ and $\theta .$ (c) Determine the magnetic moment $d \overrightarrow{\boldsymbol{\mu}}$ of the ring by multiplying $d t$ by the area enclosed by the ring. (d) Determine the magnetic moment of the spherical shell by integrating over the sphere. Express your result in terms of the total charge $Q=4 \pi R^{2} \sigma$ (c) Now consider a solid sphere of radius $R$ with volume charge density $\rho=\rho(r),$ where $r$ is the distance from the center, spinning with angular velocity $\omega$ about the $z$ -axis. Determine its magnetic moment by integrating $\overrightarrow{\boldsymbol{\mu}}=\int d \overrightarrow{\boldsymbol{\mu}},$ where $d \overrightarrow{\boldsymbol{\mu}}$ is now the magnetic moment associated with the shell at radius $r$ with differential width $d r$. Express your answer in terms of an integral $\int_{0}^{R} r^{4} \rho(r) d r .$ (f) The angular momentum of the

Name: Problem 57, Chapter 41: University Physics with Modern Physics
Uploaded: 2021-11-16T19:15:52-08:00
Duration: 8 min 31 s
Channel: Keshav Singh

   The magnetic moment $\overrightarrow{\boldsymbol{\mu}}$ of a spinning object with charge $Q$ and mass $M$ is proportional to its angular momentum $\vec{L}$ according to $\overrightarrow{\boldsymbol{\mu}}=g(Q / 2 M) \overrightarrow{\boldsymbol{L}} .$ The dimensionless coefficient $g$ is known as the $g$ -factor of the object. Consider a spherical shell with radius $R$ and uniform charge density $\sigma$ spinning with angular velocity $\omega$ around the $z$ -axis. (a) What is the differential area da of the ring at latitude $\theta,$ with width $R d \theta$, as shown in Fig. $\mathrm{P} 41.57 ?$ (b) The current $d l$ carried by the ring is its charge $\sigma$ da divided by the period of rotation. Determine $d I$ in terms of $R, \sigma, \omega,$ and $\theta .$ (c) Determine the magnetic moment $d \overrightarrow{\boldsymbol{\mu}}$ of the ring by multiplying $d t$ by the area enclosed by the ring. (d) Determine the magnetic moment of the spherical shell by integrating over the sphere. Express your result in terms of the total charge $Q=4 \pi R^{2} \sigma$ (c) Now consider a solid sphere of radius $R$ with volume charge density $\rho=\rho(r),$ where $r$ is the distance from the center, spinning with angular velocity $\omega$ about the $z$ -axis. Determine its magnetic moment by integrating $\overrightarrow{\boldsymbol{\mu}}=\int d \overrightarrow{\boldsymbol{\mu}},$ where $d \overrightarrow{\boldsymbol{\mu}}$ is now the magnetic moment associated with the shell at radius $r$ with differential width $d r$. Express your answer in terms of an integral $\int_{0}^{R} r^{4} \rho(r) d r .$ (f) The angular momentum of the

University Physics with Modern Physics

Roger A. Freedman,… 15th Edition

Chapter 41, Problem 57

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Solved by Expert Keshav Singh

11/16/2021

Step 1

This gives us: \[da = R d\theta \times 2\pi R \sin\theta = 2\pi R^2 \sin\theta d\theta.\] Show more…

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The magnetic moment $\overrightarrow{\boldsymbol{\mu}}$ of a spinning object with charge $Q$ and mass $M$ is proportional to its angular momentum $\vec{L}$ according to $\overrightarrow{\boldsymbol{\mu}}=g(Q / 2 M) \overrightarrow{\boldsymbol{L}} .$ The dimensionless coefficient $g$ is known as the $g$ -factor of the object. Consider a spherical shell with radius $R$ and uniform charge density $\sigma$ spinning with angular velocity $\omega$ around the $z$ -axis. (a) What is the differential area da of the ring at latitude $\theta,$ with width $R d \theta$, as shown in Fig. $\mathrm{P} 41.57 ?$ (b) The current $d l$ carried by the ring is its charge $\sigma$ da divided by the period of rotation. Determine $d I$ in terms of $R, \sigma, \omega,$ and $\theta .$ (c) Determine the magnetic moment $d \overrightarrow{\boldsymbol{\mu}}$ of the ring by multiplying $d t$ by the area enclosed by the ring. (d) Determine the magnetic moment of the spherical shell by integrating over the sphere. Express your result in terms of the total charge $Q=4 \pi R^{2} \sigma$ (c) Now consider a solid sphere of radius $R$ with volume charge density $\rho=\rho(r),$ where $r$ is the distance from the center, spinning with angular velocity $\omega$ about the $z$ -axis. Determine its magnetic moment by integrating $\overrightarrow{\boldsymbol{\mu}}=\int d \overrightarrow{\boldsymbol{\mu}},$ where $d \overrightarrow{\boldsymbol{\mu}}$ is now the magnetic moment associated with the shell at radius $r$ with differential width $d r$. Express your answer in terms of an integral $\int_{0}^{R} r^{4} \rho(r) d r .$ (f) The angular momentum of the

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Key Concepts

Magnetic Dipole Moment

The magnetic dipole moment is a measure of the strength and orientation of a magnetic source. For current loops, it is defined as the product of the current and the area enclosed by the loop. In the context of a spinning charged object, the distributed moving charges generate current loops, and their contributions sum to form the overall magnetic moment of the system.

Rotation-Induced Current

When a charged object rotates, its moving charges constitute an effective current. The current associated with a differential element of the object can be determined by dividing the charge passing through by the period of rotation. This concept is essential in linking a rotating charge distribution to its magnetic effects.

Gyromagnetic Ratio (g-factor)

The gyromagnetic ratio relates an object's magnetic moment to its angular momentum. Expressed with a dimensionless coefficient known as the g-factor, this ratio provides insight into how effectively angular momentum is converted into magnetic moment in various physical systems, ranging from classical spinning objects to elementary particles.

Spherical Coordinate Integration

Problems involving spherical objects often require the use of spherical coordinates to express differential elements such as area or volume. Integration over these elements accounts for the contributions from all parts of the object. This method is particularly important when computing physical quantities that are distributed over a sphere or spherical shell.

Differential Area Element in Spherical Geometry

The concept of a differential area element on a sphere is crucial for setting up integrals over curved surfaces. Typically, the area element on a sphere is expressed in terms of the spherical coordinate angles (for instance, latitude and longitude). This allows one to precisely account for the contributions from small regions on the sphere when computing quantities like total charge or magnetic moment.

Angular Momentum of Rotating Bodies

Angular momentum in a rotating body is the measure of the rotational motion, determined by the distribution of mass and the angular velocity. For charged bodies, the connection between angular momentum and magnetic moment (often mediated by the gyromagnetic ratio) is a fundamental aspect of electromagnetic theory, highlighting the interplay between mechanical rotation and electromagnetic effects.

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Transcript

00:01 For this problem on the topic of quantum mechanics, we are told that for spinning object with charge q and mass m, the magnetic moment mu is proportional to its angular momentum.

00:09 We then want to consider a spherical shell that it is r and uniform charge density sigma spinning with angular velocity omega about the x -axis and solve for various quantities of this motion.

00:23 Now, we want to calculate the magnetic moment of a spinning sphere of charge and the differential area, da, is equal to r d theta times 2 pi r sine theta d theta which becomes 2 pi r squared times the sign of theta d theta next we want to find the current di and the current di is the surface charge density sigma times da divided by t, where t is the period, and we know that t is 2 pi over omega.

01:23 So combining these with the relation with the result from part a, we get di to be sigma times 2 pi r squared, sine theta, d theta, divided by 2 pi over omega.

01:42 Simplifying we get the i to be sigma omega r squared sine theta d theta in part c if we use the results from a and b we get the differential magnetic moment d mu to be a di which is the area pi times r sine theta squared times the current element which is sigma omega r squared sine theta d theta and then this simplifies to pi sigma omega r to the power of theta sine cubed of theta d theta next for part b to find the magnetic moment we can integrate the expression above and so the magnetic moment, mu, is the integral from 0 to pi of pi sigma omega r to the 4th, sign of theta cubed d theta.

03:13 And so performing this integral, we get this to be 4 pi sigma omega r to the power of 4 divided by 3.

03:26 And so if we use sigma equal to the total charge q over the total area 4 pi r squared, means we get a magnetic moment mu that is equal to 4 pi omega r to the power of 4 divided by 3 multiplied by q over 4 pi r squared.

03:57 And so the magnetic moment simplifies to omega r squared q divided by 3.

04:14 For part e, we now have a solid sphere and consider the sphere to be a series of thin concentric shells, each of thickness dr...

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