Calculate the magnetic moment of a rotating sphere

[LeoJakob]

Homework Statement

A sphere with radius $R$ is spatially homogeneously loaded and rotates with constant angular velocity $\vec{ \omega}$ around the $z$ axis running through the center of the sphere.

Calculate the magnetic moment of the sphere $\vec m$

Relevant Equations

$\vec{m}=\frac{1}{2} \int \limits_{V} \vec{r}^{\prime} \times \vec{j}\left(\vec{r}^{\prime}\right) d V, \vec{j}(\vec{r}, t)=\rho(\vec{r}, t) \vec{v}(\vec{r}, t), \vec{v}=\vec{\omega} \times \vec{r}$

Ich wäre Ihnen sehr dankbar, wenn Sie sich meine Lösung der folgenden Übung ansehen:

A sphere with radius $R$ is spatially homogeneously loaded and rotates with constant angular velocity $\vec{ \omega}$ around the $z$ axis running through the center of the sphere.
Calculate the magnetic moment of the sphere $\vec m$

First I need to calculate $\vec{j}(\vec{r}, t)=\rho(\vec{r}, t) \vec{v}(\vec{r}, t)$

I will use spherical coordinates:

(i) I calculated $\rho (r)=\frac{3 Q}{4 \pi R^{3}} \theta(R-r)$ with the total charge $Q$
(ii)

$$\vec{\omega}=\omega \vec{e}_{z}, \vec{r}=r \vec{e}_{r}$$

$$\vec{v}=\vec{\omega} \times \vec{r}=r \omega\left(\vec{e}_{z} \times \vec{e}_{r}\right)=r \omega \vec{e}_{\phi}$$

$$\vec{j}(\vec{r})=\rho(r) \vec{v}=\rho(r) \vec{\omega} \times \vec{r}=\rho(r) r \omega \vec{e}_{\phi}=j(r) \vec{e}_{\phi} $$

$$\vec{m}=\frac{1}{2} \int \limits_{V} \vec{r}^{\prime} \times \vec{j}\left(\vec{r}^{\prime}\right) d V \\
=\frac{1}{2} \int \limits_{V}\left(r^{\prime} \vec{e}_{r'}\right) \times\left(j\left(r^{\prime}) \vec{e}_{\phi}\right) d V\right. \\
=\frac{1}{2} \int \limits_{V} r^{\prime} j\left(r^{\prime}\right)(\underbrace{\vec{e}_{r'} \times \vec{e}_{\phi}}_{\overrightarrow{e_{z}}}) d V \\
=\frac{1}{2} \int \limits_{0}^{2 \pi} d \phi \int \limits_{0}^{\pi} \sin \theta d \theta \int \limits_{0}^{\infty} r'\left(\frac{3 Q}{4 \pi R^{3}} \theta(R-r')\right) \cdot r^{\prime} \omega\left(r^{\prime}\right)^{2} d r^{\prime} \vec{e}_{z} \\
=\frac{1}{2}(2 \pi)(-2) \frac{3 Q}{4 \pi R^{3}} \int \limits_{0}^{R}\left(r^{\prime}\right)^{4} d r^{\prime} \vec{e}_{z} \\ =-\frac{3}{2} \frac{Q \omega}{R^{3}}\left[\frac{\left(r^{\prime}\right)^{5}}{5}\right]_{0}^{R}\vec{e}_{z} =-\frac{3}{10} Q \omega R^{2} \vec{e}_{z}
$$

The units are correct $[ A\cdot m^2]$

 

[TSny]

$$\vec{v}=\vec{\omega} \times \vec{r}=r \omega\left(\vec{e}_{z} \times \vec{e}_{r}\right)=r \omega \vec{e}_{\phi}$$

In spherical coordinates, $\vec {e}_z \times \vec {e}_r \neq \vec {e}_\phi$, in general.
A sketch will help.

$$\frac{1}{2} \int \limits_{V} r^{\prime} j\left(r^{\prime}\right)(\underbrace{\vec{e}_{r'} \times \vec{e}_{\phi}}_{\overrightarrow{e_{z}}}) d V $$

$\vec {e}_{r'} \times \vec {e}_{\phi} \neq \vec {e}_z$, in general.

 

[kuruman]

Homework Statement: A sphere with radius $R$ is spatially homogeneously loaded and rotates with constant angular velocity $\vec{ \omega}$ around the $z$ axis running through the center of the sphere.
(i) I calculated $\rho (r)=\frac{3 Q}{4 \pi R^{3}} \theta(R-r)$ with the total charge $Q$

What does "spatially homogeneously loaded" mean? I think it means "has uniform charge density" and that you should have $\rho = \text{const.}$ Besides, your expression does not make sense because it shows a $\theta$ dependence that is not periodic. When $\theta = 2\pi, 4\pi,\dots$, you should get the same value as when $\theta=0$.

The way I would do this problem is write $dq=\rho~dV=\rho~ r^2 dr~\sin\theta d\theta~d\phi$ and consider the magnetic moment of a current loop of radius $r$ and current $dI=\dfrac{dq}{dt}.$

Find the contribution to the magnetic moment of this current loop and add all such contributions over the volume of the sphere.

 

[TSny]

your expression does not make sense because it shows a $\theta$ dependence that is not periodic.

I think the $\theta$ in the expression $\rho (r)=\frac{3 Q}{4 \pi R^{3}} \theta(R-r)$ is meant to represent the Heaviside step function.

 

[LeoJakob]

Besides, your expression does not make sense because it shows a θ dependence that is not periodic.

I'm really sry for that, I meant the heaviside-function and should have wrote $\vartheta$ for the spherical coordinates. I will give you a new calculation in a few minutes

 

[kuruman]

I'm really sry for that, I meant the heaviside-function and should have wrote $\vartheta$ for the spherical coordinates. I will give you a new calculation in a few minutes

Why do you need the Heaviside function? The volume charge density is $\rho=\frac{3Q}{4\pi R^3}$ as you say. If you integrate over $r$ from 0 to $R$ you implicitly take into consideration that there is no charge in the region $r>R$.

 

[LeoJakob]

In spherical coordinates, $\vec {e}_z \times \vec {e}_r \neq \vec {e}_\phi$, in general.
A sketch will help.


$\vec {e}_{r'} \times \vec {e}_{\phi} \neq \vec {e}_z$, in general.

But $\vec {e}_{r'} \times \vec {e}_{\phi} = \vec {e}_{\theta}$ is correct?

Why do you need the Heaviside function? The volume charge density is $\rho=\frac{3Q}{4\pi R^3}$ as you say. If you integrate over $r$ from 0 to $R$ you implicitly take into consideration that there is no charge in the region $r>R$.

True!

A sphere with radius $R$ is spatially homogeneously loaded and rotates with constant angular velocity $\vec{ \omega}$ around the $z$ axis running through the center of the sphere.
Calculate the magnetic moment of the sphere $\vec m$

First I need to calculate $\vec{j}(\vec{r}, t)=\rho(\vec{r}, t) \vec{v}(\vec{r}, t)$

Failed attempt:
I tried to use spherical coordinates. Because of the radial symmetrie the magnetic moment will only be dependent on the distance $r$ from the origin. I can set my coordinate system so that the position vector $\vec r$ is on the z-axis. Don't know the next step....

The way I would do this problem is write $dq=\rho~dV=\rho~ r^2 dr~\sin\theta d\theta~d\phi$ and consider the magnetic moment of a current loop of radius $r$ and current $dI=\dfrac{dq}{dt}.$

I only know this formula $\vec{m}=\frac{1}{2} \int \limits_{V} \vec{r}^{\prime} \times \vec{j}\left(\vec{r}^{\prime}\right) d V$. I don’t understand what your idea is: $j=\frac{I}{S}=\frac{dq}{dt}\cdot \frac{1}{S}=\frac{d(\rho~ r^2 dr~\sin\theta d\theta~d\phi)}{dt}\cdot \frac{1}{S}$

$S$ is the cross-sectional area.

 

[TSny]

But $\vec {e}_{r'} \times \vec {e}_{\phi} = \vec {e}_{\theta}$ is correct?

Almost. Can you see that the sign is wrong? That is, you should have $\vec {e}_{r'} \times \vec {e}_{\phi} = -\vec {e}_{\theta}$.

By symmetry, the magnetic moment will be in the $z$-direction. So, you only need the $z$-component of $\vec {e}_{r'} \times \vec {e}_{\phi}$. What is the $z$-component of $-\vec{e}_{\theta}$ in terms of the angle $\theta$?

Second, you need to correct your expression for $\vec{v}$. You wrote correctly that $\vec v = \vec{\omega} \times \vec r = r \omega \vec e_z \times \vec e_r$. But then you wrote $\vec e_z \times \vec e_r= \vec e_{\phi}$, which is incorrect. Note that $\vec e_z$ and $\vec e_r$ are not perpendicular to one another. You should get that $\vec e_z \times \vec e_r$ equals some function of $\theta$ times $\vec e_{\phi}$.

If you make these two corrections in your calculation, I think you will get the correct answer. You will find that the only change will be in the integration over $\theta$.

 

[kuruman]

I only know this formula $\vec{m}=\frac{1}{2} \int \limits_{V} \vec{r}^{\prime} \times \vec{j}\left(\vec{r}^{\prime}\right) d V$. I don’t understand what your idea is: $j=\frac{I}{S}=\frac{dq}{dt}\cdot \frac{1}{S}=\frac{d(\rho~ r^2 dr~\sin\theta d\theta~d\phi)}{dt}\cdot \frac{1}{S}$

$S$ is the cross-sectional area.

I will show you my solution once you solve this problem the way you understand it best. I don't want to interfere with the alternative path that @TSny is guiding you through.

 

[LeoJakob]

By symmetry, the magnetic moment will be in the $z$-direction. So, you only need the $z$-component of $\vec {e}_{r'} \times \vec {e}_{\phi}$. What is the $z$-component of $-\vec{e}_{\theta}$ in terms of the angle $\theta$?

$(\vec {e}_{r'} \times \vec {e}_{\phi})_z=(-\vec {e}_{\theta})_z=\sin \theta$? I don't know how I can represent $\vec e_z$ in spherical coordinates?

 

[TSny]

$(\vec {e}_{r'} \times \vec {e}_{\phi})_z=(-\vec {e}_{\theta})_z=\sin \theta$?

Yes. Good.

 

[TSny]

I don't know how I can represent $\vec e_z$ in spherical coordinates?

From a sketch you should be able to identify the angle that $\vec e_z$ makes to each of the spherical-coordinate unit vectors: $\vec e_r, \vec e_{\theta}$ and $\vec e_{\phi}$. Then you can fill in the parentheses in $$\vec e_z = (\,\,\,) \vec e_r + (\,\,\,) \vec e_{\theta} + (\,\,\,)\vec e_{\phi}.$$
You should find that the content of one of the parentheses is zero and the others are functions of $\theta$.

 

[LeoJakob]

Ahhhh, $\vec e_\phi$ is zero becaues it is perpendicular to the $z$- axis., so it will not contribute to a $z$ component.
So I'm now at this point:

$$ \left(\begin{array}{l}0 \\ 0 \\ 1\end{array}\right)=\lambda\left(\begin{array}{c}\sin \theta \cos \phi \\ \sin \theta \sin \phi \\ \cos \theta\end{array}\right)+\mu\left(\begin{array}{c}\cos \theta \cos \phi \\ \cos \theta \sin \phi \\ -\sin \theta\end{array}\right) $$

It is necessary that:
$$\sin \theta \cos \phi=\sin \theta \sin \phi=\cos \theta \cos \phi=\cos \theta \sin \phi=0$$

I haven’t gotten any further with my thought process.

Edit - new thoughts: $$ \begin{array}{l}\phi=0 \\ \left(\begin{array}{l}0 \\ 0 \\ 1\end{array}\right)= \vec{e}_{z}=\cos \theta \vec e_r -\sin \theta \vec{e}_{\theta} ?\end{array} $$

 

[TSny]

Edit - new thoughts: $$ \vec{e}_{z}=\cos \theta \vec e_r -\sin \theta \vec{e}_{\theta} ?$$

Yes, that's it.

 

[LeoJakob]

$$ \begin{array}{l}m=\frac{1}{2} \int \limits_{V} \vec{r}^{\prime} \times \vec{j}\left(\vec{r}^{\prime}\right) d V \\ \vec{\omega}=\omega \overrightarrow{e_{z}}=\omega\left(\cos \theta \vec{e}_{r}-\sin \theta \overrightarrow{e_{\theta}}\right) \\ \vec{v}=\vec{\omega} \times \vec{r}=\left(\omega \vec{e}_{z}\right) \times\left(r \vec{e}_{r}\right) \\ =\omega r\left[\left(\cos \theta \vec{e}_{r}-\sin \theta \vec{e}_{\theta}\right) \times \vec{e}_{r}\right] \\ =\operatorname{\omega r}[\cos \theta(\underbrace{\left(\vec{e}_{r} \times \vec{e}_{r}\right.}_{=0})-\sin \theta(\underbrace{\overrightarrow{e_{\theta}} \times \vec{e}_{r}}_{-\vec{e}_{\phi}})] \\ =\omega r \sin \theta \vec{e}_{\phi} \\ \vec{j}=\rho \vec{v}=\rho \cdot \omega r \sin \theta \vec{e}_{\phi} \\ \vec{r}^{\prime} \times \vec{j}=\left(r^{\prime} \overrightarrow{e_{r^{\prime}}}\right) \times \vec{j} \\ =p \omega\left(r^{\prime}\right)^{2} \sin \theta(\underbrace{\overrightarrow{e_{r}} \times \overrightarrow{e_{\phi}}}_{-\overrightarrow{e_{\theta}}}) \\\end{array} $$

$$ \begin{aligned} \vec{m} & =\frac{1}{2} \int \limits_{0}^{2 \pi} d \phi \int \limits_{0}^{R} \rho \omega\left(r^{\prime}\right)^{4} d r^{\prime} \int \limits_{0}^{\pi} \sin ^{2} \theta d \theta\left(-\vec{e}_{\theta}\right) \\ & =-\rho \omega \pi\left[\frac{\left(r^{\prime}\right)^{5}}{5}\right]_{0}^{R} \cdot \frac{\pi}{2} \vec{e}_{\theta} \\ & =-p \omega \pi^{2} \frac{R^{5}}{10} \vec{e}_{\theta} \\ & =\left(\frac{3 Q}{4 \pi R^{3}}\right) \omega \pi^{2} \frac{R^{5}}{10} \overrightarrow{e_{\theta}}=-\frac{3\pi}{40}Q\omega R^2\overrightarrow{e_{\theta}}\end{aligned} $$

 

[TSny]

Everything looks good up to here:

$$ \begin{aligned} \vec{m} & =\frac{1}{2} \int \limits_{0}^{2 \pi} d \phi \int \limits_{0}^{R} \rho \omega\left(r^{\prime}\right)^{4} d r^{\prime} \int \limits_{0}^{\pi} \sin ^{2} \theta d \theta\left(-\vec{e}_{\theta}\right) \\ \end{aligned} $$

$\vec e_{\theta}$ has a direction that depends on the value of $\theta$. So, you can't treat $\vec e_{\theta}$ as a constant vector when integrating over $\theta$.

We know that $\vec m$ is in the $z$ direction. So, we only need $m_z$.$$ m_z = \vec m \cdot \vec{e}_z = \frac{1}{2} \int \limits_{0}^{2 \pi} d \phi \int \limits_{0}^{R} \rho \omega\left(r^{\prime}\right)^{4} d r^{\prime} \int \limits_{0}^{\pi} \sin ^{2} \theta \left[\left(-\vec{e}_{\theta}\right) \cdot \vec{e}_z\right] d \theta$$ Substitute for $\left(-\vec{e}_{\theta}\right) \cdot \vec{e}_z$ and then do the $\theta$ integration.

 

[LeoJakob]

We know that $\vec m$ is in the $z$ direction. So, we only need $m_z$.$$ m_z = \vec m \cdot \vec{e}_z = \frac{1}{2} \int \limits_{0}^{2 \pi} d \phi \int \limits_{0}^{R} \rho \omega\left(r^{\prime}\right)^{4} d r^{\prime} \int \limits_{0}^{\pi} \sin ^{2} \theta \left[\left(-\vec{e}_{\theta}\right) \cdot \vec{e}_z\right] d \theta$$ Substitute for $\left(-\vec{e}_{\theta}\right) \cdot \vec{e}_z$ and then do the $\theta$ integration.

$$ \begin{array}{l} m_z = \vec m \cdot \vec{e}_z = \frac{1}{2} \int \limits_{0}^{2 \pi} d \phi \int \limits_{0}^{R} \rho \omega\left(r^{\prime}\right)^{4} d r^{\prime} \int \limits_{0}^{\pi} \sin ^{2} \theta \left[\left(-\vec{e}_{\theta}\right) \cdot \vec{e}_z\right] d \theta=\frac{1}{2} 2 \pi \int \limits_{0}^{R} \rho \omega\left(r^{\prime}\right)^{4} d r^{\prime} \int \limits_{0}^{\pi} \sin ^{2} \theta(\sin \theta) d \theta \\ =p \omega \pi \frac{R^{5}}{5} \cdot \frac{4}{3}=\frac{3}{4} \frac{Q}{\pi R^{3}} \omega \pi \frac{4}{3} \frac{R^{5}}{5}=\frac{1}{5} \omega Q R^{2}\end{array} $$

Thank you very much for taking the time to help me :) and for your patience.

 

[TSny]

That looks very good. Nice work.

Here's something that you might see in a later course. For any object with uniform charge density $\rho_c$ and uniform mass density $\rho_m$ and spinning about a fixed axis, there is a simple relation between the magnetic moment $\vec M$ and the angular momentum $\vec L$. $$\vec M = \frac 1 2 \int \vec r \times j(\vec r) dV = \frac 1 2 \int \vec r \times \rho_c \vec v(\vec r) dV = \frac 1 2 \rho_c \int \vec r \times \vec v(\vec r) dV = \frac 1 2 \frac{\rho_c}{\rho_m}\int \vec r \times \vec v(\vec r) \rho_m dV$$ The integral on the far right gives the angular momentum. So, $$\vec M = \frac 1 2 \frac{\rho_c}{\rho_m} \vec L = \frac{Q}{2m} \vec L, $$ where $m$ is the total mass of the object.

For a sphere with uniform mass density, we have $\vec L = I \vec \omega = \frac 2 5 m R^2 \vec \omega$. Substituting this into the relation above, $$\vec M = \frac 1 5 Q R^2 \vec{\omega}.$$