Curl of the vector potential produced by a solenoid

[phy124]

Homework Statement

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Homework Equations

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I was looking at Example 5.12 in Griffiths (http://screencast.com/t/gGrZEPBpk0) and I can't manage to work out how to verify that the curl of the vector potential, A, is equal to the magnetic field, B.

I believe my problem lies in confusion about how to apply the equation for curl in cylindrical coordinates (given the final vector potential in equation 5.71).

The Attempt at a Solution

I was unable to make progress while trying to apply the formula for curl in cylindrical coordinates as per https://upload.wikimedia.org/math/1/d/c/1dc228ba3864e64bff1d49fd0984a80f.png

Am I using the wrong equation? If not, how should I be going about doing it? Any assistance is appreciated.

 

[BvU]

You're doing fine. Griffiths wants to avoid confusion between r, R and ##\rho## so he uses s. In the curl expression that is ##\rho##.

 

[phy124]

Ok thank you.

So then do I write:

[tex]A_{\phi} = \frac{\mu_0nIR^2}{2\rho}=\frac{\mu_0nI(\rho^2 + z^2)}{2\rho}[/tex]

Then after simplification as all other terms are zero [tex]\bigtriangledown \times A = -\frac{\partial A_{\phi}}{\partial z}\hat{\rho}+\frac{1}{\rho}\left(\frac{\partial (\rho A_{\phi})}{\partial \rho}\right)\hat{z}[/tex]

 

[BvU]

I don't think ##R^2 = \rho^2 + z^2##. I think it's the radius of the solenoid. Simplifies things !

 

[phy124]

Sorry I don't think I'm following, I'm still not sure what I'm supposed to have for [tex]A_{\phi}[/tex] then. I'm thinking that I need to get [tex]\mu_0 n I \hat{z}[/tex] out of this as the result but I've tried many different things for [tex]A_{\phi}[/tex] with no progress. What should the value of [tex]A_{\phi}[/tex] be please.

Ugh, don't know how to format the latex, sorry.

 

[BvU]

For ##A_\phi ## you have 5.70 for inside and 5.71 for outside.
e.g. 5.70 :
$$A_\phi = {\mu_0 nI\over 2} s\hat \phi $$doesn't depend on z, so:

For the curl then the only nonzero is ## \displaystyle \frac{1}{\rho}\left(\frac{\partial (\rho A_{\phi})}{\partial \rho}\right)\hat{z}## , in this case ##\displaystyle\frac{1}{s}\left(\frac{\partial (s A_{\phi})}{\partial s}\right)\hat{z}## . Bingo !

 

[phy124]

But then for the outside wouldn't that mean that [tex]\frac{1}{\rho}\left({\frac{\partial (\rho A_{\phi})}{\partial \rho}}\right)\hat{z}=\frac{1}{\rho}\left({\frac{\partial (\rho\frac{ \mu_0 n I R^2}{2 \rho})}{\partial \rho}}\right)\hat{z}=\frac{1}{\rho}\left({\frac{\partial (\frac{ \mu_0 n I R^2}{2})}{\partial \rho}}\right)\hat{z} = 0[/tex]

Which isn't equal to [tex]B = \mu_0 n I[/tex]?

 

[BvU]

And are you unhappy about that ? You just found the same result as 5.57 in example 5.9 ! Bravo !

 

[phy124]

Oh... *sigh*

Thank you for your assistance BvU, it was much appreciated!

 

[BvU]

You're welcome. I learn too (haven't worked with vector potential since university).