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Total moment density of electromagnetic fields

  1. May 27, 2017 #1
    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)

    Thank you in advance!!
     
  2. jcsd
  3. May 28, 2017 #2

    TSny

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    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.
     
  4. May 28, 2017 #3
    Hello TSny!!
    Thank you for your comments.
    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##:smile::smile::smile:

    Thank you very much for your help!!
     
  5. May 28, 2017 #4

    TSny

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    OK. I overlooked the ##c^2##.:oops:

    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.
    upload_2017-5-28_18-36-32.png

    Yes. Good!
     
  6. May 29, 2017 #5
    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!!:smile::smile::smile:
     
  7. May 29, 2017 #6

    TSny

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    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.
     
  8. May 30, 2017 #7
    Thank you. You have helped me a lot!!
     
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