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

In summary, the vector potential in cylindrical coordinates is different from the vector potential in spherical coordinates.
  • #1
Adesh
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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
Screen Shot 2020-05-09 at 2.36.31 PM.png
.

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.
 
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  • #2
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
$$\vec{A}=0\hat r+\frac{\mu_0 nI r\sin\theta}{2}\hat \phi +0\hat\theta$$
 
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  • #3
if you work in cylindrical coordinates you ll get the result very fast.
 
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  • #4
@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.
 
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Related to I'm not getting the curl of vector potential equal to magnetic field

1. Why is the curl of vector potential not equal to the magnetic field?

The curl of vector potential is not equal to the magnetic field because the vector potential is a mathematical construct used to simplify the equations of electromagnetism. It is not a physical quantity, but rather a mathematical tool that helps us understand and calculate the behavior of magnetic fields.

2. Can the curl of vector potential ever be equal to the magnetic field?

No, the curl of vector potential can never be equal to the magnetic field because they represent different physical quantities. The magnetic field is a physical field that can be measured and has real-world effects, while the vector potential is a mathematical construct used to simplify calculations.

3. How is the vector potential related to the magnetic field?

The vector potential is related to the magnetic field through the curl operator. The curl of the vector potential gives us the magnetic field, just as taking the derivative of a function gives us its slope. However, the vector potential itself is not the magnetic field, but rather a mathematical representation of it.

4. Why do we use the vector potential if it is not equal to the magnetic field?

We use the vector potential because it simplifies the equations of electromagnetism and makes them easier to solve. It also allows us to describe the behavior of magnetic fields in a more elegant and concise way. Additionally, the vector potential has other important applications in quantum mechanics and relativity.

5. Are there any other quantities that are similar to the vector potential?

Yes, there are other quantities that are similar to the vector potential, such as the electric potential and the gravitational potential. These are all mathematical constructs that help us understand and calculate the behavior of physical fields. However, it is important to remember that they are not physical quantities themselves, but rather tools that aid in our understanding of the physical world.

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