# Curl of the vector potential produced by a solenoid

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1. Aug 27, 2015

### phy124

1. The problem statement, all variables and given/known data / 2. Relevant equations
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).

3. 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.

2. Aug 27, 2015

### 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 $\rho$.

3. Aug 27, 2015

### phy124

Ok thank you.

So then do I write:

$$A_{\phi} = \frac{\mu_0nIR^2}{2\rho}=\frac{\mu_0nI(\rho^2 + z^2)}{2\rho}$$

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

Last edited: Aug 27, 2015
4. Aug 27, 2015

### BvU

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

5. Aug 27, 2015

### phy124

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

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

Last edited: Aug 27, 2015
6. Aug 27, 2015

### 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 !

7. Aug 27, 2015

### phy124

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

Which isn't equal to $$B = \mu_0 n I$$?

8. Aug 27, 2015

### BvU

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

9. Aug 27, 2015

### phy124

Oh... *sigh*

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

10. Aug 27, 2015

### BvU

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