Find angular momentum of EM field in terms of q and ##\Phi##

[WeiShan Ng]

Homework Statement

A point charge q sits at the origin. A magnetic field $\mathbf{B} (\mathbf{r})=B(x,y)\mathbf{\hat{z}}$ fills all of space. The problem asks us to write down an expression for the total electromagnetic field angular momentum $\bf{L_{EM}}$, in terms of q and the magnetic flux, $\Phi_B$ through the xy-plane

Homework Equations

momentum density
$$\mathbf{g} = \epsilon_0 (\mathbf{E} \times \mathbf{B})$$
angular momentum density:
$$l = \vec{r} \times \vec{g} = \epsilon_0 [\vec{r}\times (\vec{E}\times \vec{B})]$$

The Attempt at a Solution

The electric field from q is
$$\vec{E}=\frac{Q}{4\pi \epsilon_0 r^2} \mathbf{\hat{r}}$$
The magnetic field is
$$\vec{B}=B(x,y) \mathbf{\hat{z}}$$
So the angular momentum density is
$$\begin{align*} \mathbf{l} &= \epsilon_0 [\vec{r} \times (\mathbf{E}\times \mathbf{B})] \\
&=\epsilon_0 r\frac{QB}{4\pi \epsilon_0 r^2} [\mathbf{\hat{r} \times \hat{r} \times \mathbf{\hat{z}}}] \end{align*} $$
and
$$[\mathbf{\hat{r}\times \hat{r} \times \hat{z}}] = -r \sin \theta \cos \theta \cos \phi \mathbf{\hat{x}}+ r \sin \theta \cos \theta \sin \phi \mathbf{\hat{y}} + -r \sin^2 \theta \mathbf{\hat{z}}$$
The total angular momentum will be
$$\mathbf{L} = \iiint \mathbf{l} \, r^2 \sin \theta \, dr d\theta d\phi$$
Since we know that $\int \sin \phi \, d\phi$ and $\int \cos \phi \, d\phi$ equal to zero in the interval $[0, 2\pi]$, we can disregard x and y component, so the total angular momentum will be
$$\begin{align*} \mathbf{L} &= \iiint \frac{QB}{4\pi r} (-r \sin^2 \theta ) \, r^2 \sin \theta \, dr d\theta d\phi \, \mathbf{\hat{z}} \\ &= \iiint -\frac{QBr^2}{4\pi} \sin^3 \theta \, dr d\theta d\phi \end{align*}$$
Since we are considering xy-plane, (The surface element in a surface of polar angle θ constant) the magnetic flux can be written as
$$d\Phi = BdA = B r \sin \theta d\phi dr$$
$$\Phi = \iint B r \sin \theta d\phi dr$$
My problem is how can I substitute $\Phi$ into the equation of $\mathbf{L}$, since I will have an extra $r \sin^2 \theta$ in my integral ?
$$\mathbf{L} = \iiint -(B r \sin \theta) \frac{Qr\sin^2 \theta}{4\pi} \, dr d\theta d\phi \, \mathbf{\hat{z}}$$

 

[WeiShan Ng]

Alright I found my own errors, turns out $[\mathbf{\hat{r}\times \hat{r}\times \hat{z}}] = -\sin \theta \cos \theta \cos \phi \mathbf{\hat{x}}+ \sin \theta \cos \theta \sin \phi \mathbf{\hat{y}} + - \sin^2 \theta \mathbf{\hat{z}}$ and since $\theta = \pi/2$ in xy plane, I can write $$\Phi = \iint B r d\phi dr$$ ...

 

[TSny]

So the angular momentum density is
$$\begin{align*} \mathbf{l} &= \epsilon_0 [\vec{r} \times (\mathbf{E}\times \mathbf{B})] \\
&=\epsilon_0 r\frac{QB}{4\pi \epsilon_0 r^2} [\mathbf{\hat{r} \times \hat{r} \times \mathbf{\hat{z}}}] \end{align*} $$

OK, except the triple cross product in the last line needs parentheses: $[\mathbf{\hat{r} \times \left(\hat{r} \times \mathbf{\hat{z}}\right)}]$

and
$$[\mathbf{\hat{r}\times \hat{r} \times \hat{z}}] = -r \sin \theta \cos \theta \cos \phi \mathbf{\hat{x}}+ r \sin \theta \cos \theta \sin \phi \mathbf{\hat{y}} + -r \sin^2 \theta \mathbf{\hat{z}}$$

I believe the first term on the right should not have the negative sign. Also, the factor of $r$ in each term should not be there (as you noted in your post #2).

The total angular momentum will be
$$\mathbf{L} = \iiint \mathbf{l} \, r^2 \sin \theta \, dr d\theta d\phi$$
Since we know that $\int \sin \phi \, d\phi$ and $\int \cos \phi \, d\phi$ equal to zero in the interval $[0, 2\pi]$, we can disregard x and y component

This is not the reason that the x and y components can be neglected. Note that $B(x,y)$ is an arbitrary function of $x$ and $y$. So, in spherical coordinates $B$ would be some function of $r, \theta$ and $\phi$. Thus, instead of $\int \sin \phi \, d\phi$, you would need to consider $\int B \sin \phi \, d\phi$ and this is not necessarily zero. Similarly, $\int B \cos \phi \, d\phi$ is not necessarily zero. Try to find the correct reason why the the x and y components can be neglected. Hint: Consider the magnitude and direction of the angular momentum density $\mathbf l$ at points $(x, y, z)$ and $(x, y, -z)$.

Since we are considering xy-plane, ...

You are not just considering the xy-plane. You must consider the integration to be over all of 3D space and show that this integration can be expressed in terms of $Q$ and the total magnetic flux $\Phi$ through the xy-plane.

 

[WeiShan Ng]

I try to walk through the steps to see where I have done wrong:

ry to find the correct reason why the the x and y components can be neglected. Hint: Consider the magnitude and direction of the angular momentum density ll\mathbf l at points (x,y,z)(x,y,z)(x, y, z) and (x,y,−z)(x,y,−z)(x, y, -z).

Writing the cross products in Cartesian coordinates I have $$[\mathbf{\hat{r}\times (\hat{r}\times \hat{z})}] = \frac{xz}{r^2} \mathbf{\hat{x}} + \frac{yz}{r^2} \mathbf{\hat{y}}+\frac{x^2-y^2}{r^2} \mathbf{\hat{z}}$$ where $r^2 = x^2+y^2+z^2$ So the angular momentum density, $\mathbf{l}$ will be
$$\begin{align*}\mathbf{l} &= \frac{QB(x,y)}{4\pi r^3 }(xz \mathbf{\hat{x}}+ yz\mathbf{\hat{y}}+(x^2-y^2)\mathbf{\hat{z}}) \\ &= \frac{QB(x,y)}{4\pi (x^2+y^2+z^2)^{3/2} }(xz \mathbf{\hat{x}}+ yz\mathbf{\hat{y}}+(x^2-y^2)\mathbf{\hat{z}}) \end{align*}$$
x-component and y-component of $\mathbf{l}$ will have the same magnitude but opposite direction at $\mathbf{l}(x,y,z)$ and $\mathbf{l}(x,y,-z)$ for $\forall (x,y)$. If we integrate it over all volume, they will cancel out and we only left with z-component.

You are not just considering the xy-plane. You must consider the integration to be over all of 3D space and show that this integration can be expressed in terms of QQQ and the total magnetic flux ΦΦ\Phi through the xy-plane.

The $\Phi_B$ is across xy-plane at z = 0, so it can be written as
$$d\Phi = BdA = B r \sin \theta d\phi dr$$ $$\Phi(\theta) = \iint B(r,\theta,\phi) r \sin \theta d\phi dr = \int_{-\infty}^\infty \int^{2\pi}_0 B(r,\theta,\phi) r d\phi dr $$ since $\theta = \pi/2$ at z=0
Integrate $\mathbf{l}$ over all space
$$\begin{align*}\mathbf{L} &= \iiint_{all space} \mathbf{l} \, r^2 \sin \theta d\phi dr d\theta \\
&= \int^{\pi}_0 \int^\infty_{-\infty} \int^{2\pi}_0 \frac{-QB \sin^2 \theta}{4\pi r} \mathbf{\hat{z}} \, r^2 \sin \theta d\phi dr d\theta \\
&=\int_0^{\pi} \left[ \int_{-\infty}^{\infty} \int_0^\pi \frac{-QB(r,\theta,\phi)r}{4\pi} d\phi dr \right] \sin^3 \theta d\theta \mathbf{\hat{z}} \\
&= \int_0^{\pi} \frac{-Q\Phi(\theta)}{4\pi} \sin^3 \theta d\theta \mathbf{\hat{z}} \end{align*}$$
And this will be my expression for $\mathbf{L}$. Hopefully I get it right this time...

 

[TSny]

Writing the cross products in Cartesian coordinates I have $$[\mathbf{\hat{r}\times (\hat{r}\times \hat{z})}] = \frac{xz}{r^2} \mathbf{\hat{x}} + \frac{yz}{r^2} \mathbf{\hat{y}}+\frac{x^2-y^2}{r^2} \mathbf{\hat{z}}$$ where $r^2 = x^2+y^2+z^2$

There is a sign error in the z-component term. But this will not affect your argumment for showing that the x and y components of $\mathbf{L}_z$ are zero.

So the angular momentum density, $\mathbf{l}$ will be
$$\begin{align*}\mathbf{l} &= \frac{QB(x,y)}{4\pi r^3 }(xz \mathbf{\hat{x}}+ yz\mathbf{\hat{y}}+(x^2-y^2)\mathbf{\hat{z}}) \\ &= \frac{QB(x,y)}{4\pi (x^2+y^2+z^2)^{3/2} }(xz \mathbf{\hat{x}}+ yz\mathbf{\hat{y}}+(x^2-y^2)\mathbf{\hat{z}}) \end{align*}$$
x-component and y-component of $\mathbf{l}$ will have the same magnitude but opposite direction at $\mathbf{l}(x,y,z)$ and $\mathbf{l}(x,y,-z)$ for $\forall (x,y)$. If we integrate it over all volume, they will cancel out and we only left with z-component.

Yes. Nice.

The $\Phi_B$ is across xy-plane at z = 0, so it can be written as
$$d\Phi = BdA = B r \sin \theta d\phi dr$$

This is only valid for $\theta = \pi/2$

$$\Phi(\theta) = \iint B(r,\theta,\phi) r \sin \theta d\phi dr $$

This will be the flux through the xy-plane only when $\theta = \pi/2$. I don't think spherical coordinates are going to be convenient for this problem. You might try sticking with Cartesian coordinates.

 

[WeiShan Ng]

There is a sign error in the z-component term. But this will not affect your argumment for showing that the x and y components of $\mathbf{L}_z$ are zero.
Yes. Nice.

This is only valid for $\theta = \pi/2$

This will be the flux through the xy-plane only when $\theta = \pi/2$. I don't think spherical coordinates are going to be convenient for this problem. You might try sticking with Cartesian coordinates.

Thanks for your help! I will try to redo the question in Cartesian coordinates.