I'm not getting the curl of vector potential equal to magnetic field

[Adesh]

Homework Statement

Verify that the curl of vector potential equal to the magnetic field in the case of soelnoid.

Relevant Equations

N/A

In this image of Introduction to Electrodynamics by Griffiths

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we have calculated the vector potential as $\mathbf A = \frac{\mu_0 ~n~I}{2}s \hat{\phi}$. I tried taking its curl but didn't get $\mathbf B = \mu_0~n~I \hat{z}$. In this thread, I have calculated it like this :

$$(curl~\mathbf A)_r = \frac{1}{r\sin\theta} \left[ \frac{\partial}{\partial \theta} (\sin\theta ~A_{\phi}) -\frac{\partial A_{\theta}}{\partial \phi}\right]$$
$$(curl~\mathbf A)_r = \frac{\mu_0 ~n~I}{2} \cot\theta$$

$$
(curl~\mathbf A)_{\theta}= \frac{1}{r}
\left[
\frac{1}{\sin\theta}\frac{\partial A_r}{\partial \phi} - \frac{\partial}{\partial r} (r A_{\phi}) \right]$$

$$(curl~\mathbf A)_{\theta} = \mu_0~n~I$$

$$(curl~\mathbf A)_{\phi}= \frac{1}{r} \left[ \frac{\partial}{\partial r} (r A_{\theta}) - \frac{\partial A_r}{\partial \theta} \right]$$
$$(curl~\mathbf A)_{\phi} = 0$$

Now, let's convert it into cartesian system: $$ \mathbf B = \mu_0~n~I (\frac{\cot \theta }{2} \hat r + \hat{\theta})$$
$$\frac{\cot \theta}{2} \hat r = \frac{\cos \theta \cos \phi}{2} \hat x + \frac{\cos \theta \sin \phi}{2}\hat y + \frac{\cos^2 \theta}{2} \hat z $$
$$\hat \theta= \cos\theta \cos \phi \hat x +\cos \theta \sin \phi \hat y + -\sin \theta \hat z$$
And you see I won't get the desired field if I add them component -wise.

Please guide me.

 

[Delta2]

The s in the formula makes all the difference here , it is not the $r$-coordinate of a spherical coordinate system but instead it is the $\rho$ coordinate in cylindrical coordinate system. We have to do all the calculations for $curl A$ from the start and work in cylindrical (not spherical) coordinate system.

The vector potential A in cylindrical coordinates is now $$\vec{A}=0\hat\rho+\frac{\mu_0 nI \rho}{2}\hat\phi+0\hat z$$ and it is not equal to the vector A of the previous thread, the presence of the $\rho$ coordinate makes it all different now.

Or we can still work in spherical coordinates (not recommended for this problem of infinitely long solenoid) but you first have to convert the above vector A to spherical coordinates which is easy though, we just replace $$\rho=r\sin\theta$$ since the azimuthal unit vectors $\hat \phi$ are the same in cylindrical and spherical coordinates. So the vector A in spherical coordinates is
$$\vec{A}=0\hat r+\frac{\mu_0 nI r\sin\theta}{2}\hat \phi +0\hat\theta$$

 

[Delta2]

if you work in cylindrical coordinates you ll get the result very fast.

 

[Adesh]

@Delta2 Thank you so much. I calculated it using the formula for curl in cylinderical and got the expected result.

Thank you for everything you did for me today. Your re-arrival is very beneficial to me.