- #1

Adesh

- 735

- 188

- Homework Statement:
- Find the curl of ##\mathbf A##.

- Relevant Equations:
- In the main body.

I have a vector field which is originallly written as $$ \mathbf A = \frac{\mu_0~n~I~r}{2} ~\hat \phi$$ and I translated it like this $$\mathbf A = 0 ~\hat{r},~~ \frac{\mu_0 ~n~I~r}{2} ~\hat{\phi} , ~~0 ~\hat{\theta}$$

(##r## is the distance from origin, ##\phi## is azimuthal angle and ##\theta## is the polar angle).

I want to calculate its curl, so I have two options either change it into Cartesian system and then take the curl or take the curl in spherical coordinates. Let’s take the curl in spherical coordinates:

$$(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]$$

I really don't know what ##r## and ##\sin\theta## represent as my actual vector field doesn't involve ##\sin\theta## but then also we have $$(curl~\mathbf A)_r = \frac{\mu_0 ~n~I}{2} \cot\theta$$

$$(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]$$

Since, ##A_{\theta} = A_{r}= 0## we have $$ (curl~\mathbf A)_{\phi} = 0$$

Now, I want to know if I'm correct.

How can we convert the above result into cartesian system? Please guide me through it.

(##r## is the distance from origin, ##\phi## is azimuthal angle and ##\theta## is the polar angle).

I want to calculate its curl, so I have two options either change it into Cartesian system and then take the curl or take the curl in spherical coordinates. Let’s take the curl in spherical coordinates:

$$(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]$$

I really don't know what ##r## and ##\sin\theta## represent as my actual vector field doesn't involve ##\sin\theta## but then also we have $$(curl~\mathbf A)_r = \frac{\mu_0 ~n~I}{2} \cot\theta$$

$$(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]$$

Since, ##A_{\theta} = A_{r}= 0## we have $$ (curl~\mathbf A)_{\phi} = 0$$

Now, I want to know if I'm correct.

How can we convert the above result into cartesian system? Please guide me through it.

Last edited: