Undergrad Ampere's Law For Static Magnetic Field

[BlackMelon]

Hi there!

Please refer to the picture below. I would like to understand the equation Curl(H) = J, where H is the magnetic field intensity and J is the current density. So, I inspect a simple problem.
There is a wire carrying current I in the z-axis direction. a_r, a_phi, and a_z are the unit vectors in the directions of the radius, the tangential line, and z-axis, respectively.

So, from H = I/(2*pi*r)a_phi. I take the curl of this vector (in cylindrical coordinate) and got 0. How does this relate to the current density?

Best
BlackMelon

 

[Ibix]

The field you've written down is the field outside a current carrying wire. What would you expect for the current density outside a wire?

 

[BlackMelon]

The field you've written down is the field outside a current carrying wire. What would you expect for the current density outside a wire?

Oh well it's zero. That was the silly of me LOL. Thank you very much.

By the way, I have analyzed the inside of the wire H = I*r/(2*pi*R^2) a_phi.
where r is the radius from the center of the wire to the point of interest. R is the radius of the wire. And got the correct answer:
J = curl (H) = I/(pi*R^2) a_z

 

[BlackMelon]

Adding more information to my previous comment, here is how I calculate the current density inside the wire using curl(H) = J

 

[vanhees71]

The solution in local form is indeed given by using the magnetostatic Maxwell equations as follows:
$$\vec{\nabla} \cdot \vec{B}=0, \quad \vec{\nabla} \times \vec{B}=\mu_0 \vec{j}.$$
We have given
$$\vec{j}=\begin{cases}\frac{I}{\pi R^2} \vec{e}_z &\text{for} \quad \rho \leq R \\ 0 &\text{for} \quad \rho>R. \end{cases}$$
To solve this equations it's most simple to make the ansatz
$$\vec{B}=B(\rho) \vec{e}_{\varphi}.$$
Using the formula for the curl in cylinder coordinates you get
$$\vec{\nabla} \times \vec{B}=\frac{1}{\rho} \partial_{\rho} (\rho B) \vec{e}_z.$$
From this you get for $\rho<R:$
$$\partial_{\rho} (\rho B)=\frac{\mu_0 I}{\pi R^2} \rho.$$
This can be immediately integrated to
$$B(\rho)=\frac{\mu_0 I}{2 \pi R^2}\rho + \frac{C}{\rho},$$
where $C$ is an integration constant. Since there's no singularity at $\rho=0$, you get $C=0$, i.e.,
$$B(\rho)=\frac{\mu_0 I}{2 \pi R^2}\rho \quad \text{for} \quad \rho<R.$$
For $\rho \geq R$ you have
$$\partial_{\rho} (\rho B)=0 \; \rightarrow \; B=\frac{B_0}{\rho} \quad \text{with} \quad B_0=\text{const}.$$
Now, at $\rho=R$, $B$ must be continuous, which gives
$$B(\rho)=\frac{\mu_0 I}{2 \pi \rho} \quad \text{for} \quad \rho \geq R.$$
One should also check that $\vec{\nabla} \cdot \vec{B}=0$, which however is already seen easily to be fufilled by the initial general ansatz.

 

[BlackMelon]

The solution in local form is indeed given by using the magnetostatic Maxwell equations as follows:
$$\vec{\nabla} \cdot \vec{B}=0, \quad \vec{\nabla} \times \vec{B}=\mu_0 \vec{j}.$$
We have given
$$\vec{j}=\begin{cases}\frac{I}{\pi R^2} \vec{e}_z &\text{for} \quad \rho \leq R \\ 0 &\text{for} \quad \rho>R. \end{cases}$$
To solve this equations it's most simple to make the ansatz
$$\vec{B}=B(\rho) \vec{e}_{\varphi}.$$
Using the formula for the curl in cylinder coordinates you get
$$\vec{\nabla} \times \vec{B}=\frac{1}{\rho} \partial_{\rho} (\rho B) \vec{e}_z.$$
From this you get for $\rho<R:$
$$\partial_{\rho} (\rho B)=\frac{\mu_0 I}{\pi R^2} \rho.$$
This can be immediately integrated to
$$B(\rho)=\frac{\mu_0 I}{2 \pi R^2}\rho + \frac{C}{\rho},$$
where $C$ is an integration constant. Since there's no singularity at $\rho=0$, you get $C=0$, i.e.,
$$B(\rho)=\frac{\mu_0 I}{2 \pi R^2}\rho \quad \text{for} \quad \rho<R.$$
For $\rho \geq R$ you have
$$\partial_{\rho} (\rho B)=0 \; \rightarrow \; B=\frac{B_0}{\rho} \quad \text{with} \quad B_0=\text{const}.$$
Now, at $\rho=R$, $B$ must be continuous, which gives
$$B(\rho)=\frac{\mu_0 I}{2 \pi \rho} \quad \text{for} \quad \rho \geq R.$$
One should also check that $\vec{\nabla} \cdot \vec{B}=0$, which however is already seen easily to be fufilled by the initial general ansatz.

Thank you very much for the explanation :)