I Ampere's Law For Static Magnetic Field

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    Ampere's law Curl
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The discussion centers on understanding Ampere's Law for static magnetic fields, specifically the equation Curl(H) = J, where H is the magnetic field intensity and J is the current density. The user initially misinterprets the magnetic field outside a current-carrying wire, realizing that the current density outside the wire is zero. They then correctly analyze the magnetic field inside the wire and derive the current density using Curl(H), confirming the relationship with the magnetostatic Maxwell equations. The calculations show that the magnetic field inside the wire varies linearly with the radius, while outside it follows an inverse relationship with distance. The conversation concludes with appreciation for the clarification provided.
BlackMelon
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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

1695964252946.jpeg
 
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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?
 
Ibix said:
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

1695975871845.png
 
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Likes Dale and Ibix
Adding more information to my previous comment, here is how I calculate the current density inside the wire using curl(H) = J
1696130017446.png
 
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.
 
vanhees71 said:
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 :)
 
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