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Magnetic field above loop

  1. Mar 30, 2014 #1
    1. The problem statement, all variables and given/known data
    Calculate the magnetic field above a loop shown in picture with radius ##R## and current ##I##.


    2. Relevant equations



    3. The attempt at a solution

    Firstly, curved part:

    ##\vec{H}(\vec{r})=\frac{I}{4\pi }\int \frac{d\vec{{r}'}\times (\vec{r}-\vec{{r}'})}{|\vec{r}-\vec{{r}'}|^3}cos\varphi =\frac{I}{4\pi }\int \frac{Rd\varphi \hat{e}_{\varphi }\times (\vec{r}-\vec{{r}'})}{(R^2+z^2)^{3/2}}\frac{R}{\sqrt{R^2+z^2}}##

    Where ##\hat{e}_{\varphi }\times (\vec{r}-\vec{{r}'})=|\hat{e}_{\varphi }||\vec{r}-\vec{{r}'}|sin\varphi \hat{n}=\sqrt{R^2+z^2}\hat{n}## and finally:

    ##\vec{H}(\vec{r})=\frac{I}{4\pi }\int_{\pi }^{2\pi }\frac{R^2\hat{n}d\varphi }{(R^2+z^2)^{3/2}}=\frac{IR^2}{4(R^2+z^2)^{3/2}}\hat{n}##.

    For the straight line I have some troubles with the integral...

    ##\vec{H}(\vec{r})=\frac{I}{4\pi }\int \frac{d\vec{{r}'}\times (\vec{r}-\vec{{r}'})}{|\vec{r}-\vec{{r}'}|^3}##

    Now ##d\vec{{r}'}\times (\vec{r}-\vec{{r}'})=d\vec{{r}'}\times \vec{r}## since ##d\vec{{r}'}## and ##\vec{{r}'}## are parallel. Also ##d\vec{{r}'}=dx\hat{e}_x##.

    ##\vec{H}(\vec{r})=\frac{I}{4\pi }\int_{-R}^{R}\frac{dx(\hat{e}_x\times \vec{r})}{((\vec{r}-\hat{e}_xx)(\vec{r}-\hat{e}_xx))^{3/2}}=\frac{I(\hat{e}_x\times \vec{r})}{4\pi }\int_{-R}^{R}\frac{dx}{(r^2-2\hat{e}_x\vec{r} x+x^2)^{3/2}}##

    Now what? o_O Is this even ok?
     

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    Last edited: Mar 30, 2014
  2. jcsd
  3. Mar 30, 2014 #2

    Simon Bridge

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    Presumably you've already done it, or seen it done, for an infinite straight wire?
     
  4. Mar 30, 2014 #3
    Aha, ok, I see it now...

    ##\hat{e}_x\vec{r}=0## since they are always perpendicular. (My question here: What if they werent?)

    In that case, things simplify a lot and the integral over the straight line should be ##\vec{H}(\vec{r})=\frac{IR}{2\pi r\sqrt{R^2+r^2}}\hat{e}_y##, where ##\vec{r} ## is the distance from the center of the straight line to point ##T##.
     
  5. Mar 30, 2014 #4

    Simon Bridge

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    When you don't have easy symmetry, the calculation can get arbitrarily difficult.
    In many cases the integral has to be solved numerically.

    That is why you will only see simple geometries at this stage.
    Also consider what happens to the field close to one corner of the D ;)
     
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