- #1

Adesh

- 735

- 191

- 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

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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.

**Introduction to Electrodynamics**by Griffithswe 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.