# Total moment density of electromagnetic fields

1. May 27, 2017

### 987oscar

1. The problem statement, all variables and given/known data
These is a problem from my textbook. You have an infinite solenoid with n turns per unit longitude, radius "a" and a stationary current I. In the axis there exits a uniform line charge with lineal density λ. Compute total electromagnetic momentum (lineal and angular).

2. Relevant equations
Poynting vector: $$\mathbf{S}=\frac{1}{\mu_0}\mathbf{E}\times\mathbf{B}$$.
Momentum density: $$\mathbf{g}=\mu_0\epsilon_0\mathbf{S}$$

3. The attempt at a solution
Well, I computed electric and magnetic fields as usual.
Electric field :
$$\mathbf{E}=\frac{\lambda}{2\pi\epsilon_0 r}\mathbf{\hat{r}}$$
Magnetic field (only exits inside solenoid):
$$\mathbf{B}=\mu_0 nI\mathbf{\hat{z}}$$
From here I find Poynting vector:
$$\mathbf{S}=\frac{-\lambda n I}{2\pi\epsilon_0 r}\mathbf{\hat{\phi}}$$
and momentum density:
$$\mathbf{g}=-\frac{\lambda n I}{2\pi\epsilon_0 c^2}\frac{1}{r}\mathbf{\hat{\phi}}$$
From here I have angular momentum $\mathbf{l}=\mathbf{r}\times\mathbf{g}$:
$$\mathbf{l}=-\frac{\lambda nI}{2\pi\epsilon_0 c^2}\mathbf{\hat{z}}.$$
Integrating inside the solenoid for length 1 i find total angular momentum per unit longitude:
$$\mathbf{L}=\frac{-\lambda n Ia^2}{2\epsilon_0 c^2}\mathbf{\hat{z}}$$
When I try to compute total momentum per unit longitude I find:
$$\mathbf{G}=-\frac{\lambda n I}{2\pi\epsilon_0 c^2}\int_0^1dz\int_0^{2\pi}d\phi\int_0^a\frac{1}{r}rdr\mathbf{\hat{\phi}}=\frac{-\lambda nIa}{\epsilon_0 c^2}\mathbf{\hat{\phi}}$$.
Now the solution provided is $\mathbf{G}=0$ because momentum $\mathbf{g}$ forms closed lines ¿? Where is the mistake in my calculation?. I have been thinking about it but I can't find. Any idea? ($\mathbf{L}$ calculation seems to be correct)

2. May 28, 2017

### TSny

Looks like you forgot the factor $\mu_0 \epsilon_0$ in going from S to g.
Here, I think you have notational confusion. Angular momentum is calculated relative to the origin. You can let $\mathbf{R}$ be the position vector of a point relative to the origin. So, angular momentum density would be $\mathbf{l}=\mathbf{R}\times\mathbf{g}$. Be sure to distinguish between $\mathbf{R}$ and $\mathbf{r}$. You can write $\mathbf{R}$ in terms of $\mathbf{r}$, $\hat{k}$, and $z$.
In your integration, you treated $\mathbf{\hat{\phi}}$ as a constant unit vector. Think about why this is not correct.

3. May 28, 2017

### 987oscar

Hello TSny!!
I used $\mu_0\epsilon_0=\frac{1}{c^2}$ for the $\mathbf{g}$ vector.

I used cylindrical coordinates as we have an infinite solenoid. I think angular momentum is relative to Z axis and so $\mathbf{R}=\mathbf{r}$. I followed example 8.4 in Griffiths Introduction to Electrodynamics 4º Ed. He calls "s" "horizontal" distance from a point to Z axis and he makes a product independent of s. I don't know if these is correct for this problem.

Of course!! I can't treat $\mathbf{\hat{\phi}}$ as constant. I found $\mathbf{\hat{\phi}}=-\sin\phi\mathbf{\hat{x}}+\cos\phi\mathbf{\hat{y}}$ and for angular integral I now have:
$$\int_0^{2\pi}-\sin\phi\mathbf{\hat{x}}d\phi+\int_0^{2\pi}\cos\phi\mathbf{\hat{y}}d\phi=0$$
and $\mathbf{G}=0$

Thank you very much for your help!!

4. May 28, 2017

### TSny

OK. I overlooked the $c^2$.

Looks like maybe Griffiths is using $s$ for your $r$. For example, you wrote the electric field in terms of $r$ which implies that you are using $r$ for the perpendicular distance from the axis. But, when calculating the angular momentum density $\mathbf{l}$, you need to use the vector from the origin to the point where you are calculating $\mathbf{l}$. So, you need a different symbol for this vector, say $\mathbf{R}$. For a general point inside the solenoid, you should find that $\mathbf{l}$ has both an $\hat{r}$ component as well as a $\hat{z}$ component.

Yes. Good!

5. May 29, 2017

### 987oscar

OK. I think finally I've understood it with your drawing.

I make $\mathbf{R}=\mathbf{r}+\mathbf{z}=r\cos\phi\mathbf{\hat{x}}+r\sin\phi\mathbf{\hat{y}}+z\mathbf{\hat{z}}$.

Making $C=\frac{-\mu_0nI\lambda}{2\pi}$, I find $\mathbf{g}=\frac{-C}{r}\sin\phi\mathbf{\hat{x}}+\frac{C}{r}\cos\phi\mathbf{\hat{y}}$. And now $\mathbf{l}=\mathbf{R}\times\mathbf{g}=-\frac{Cz}{r}\cos\phi\mathbf{\hat{x}}-\frac{Cz}{r}\sin\phi\mathbf{\hat{y}}+C\mathbf{\hat{z}}$.

We compute total angular momentum per unit length
$$\mathbf{L}=\int_0^1dz\int_0^ardr\int_0^{2\pi}\left(-\frac{Cz}{r}\cos\phi\mathbf{\hat{x}}-\frac{Cz}{r}\sin\phi\mathbf{\hat{y}}+C\mathbf{\hat{z}}\right)d\phi.$$ But, $\int_0^{2\pi}\cos\phi\, d\phi=0$, and $\int_0^{2\pi}\sin\phi \,d\phi=0$.
Finally we get:
$$\mathbf{L}=C\pi a^2\mathbf{\hat{z}}=-\frac{\mu_0 nI\lambda a^2}{2}\mathbf{\hat{z}}.$$

Thanks a lot for your help!!

6. May 29, 2017

### TSny

Yes, that looks very good.

You can shorten the work a little by using symmetry. It should be clear that the total angular momentum $\mathbf{L}$ only has a $z$ component. So, you only need the $z$ component of the angular momentum density $\mathbf{l}$, which is what you actually calculated in your first post. So, you got the correct total angular momentum in your first post.

7. May 30, 2017

### 987oscar

Thank you. You have helped me a lot!!