# Magnetic Potential of Long Straight Wire: 65 Chars

• latentcorpse
In summary, we discussed the problem of finding the vector potential inside a long straight wire with uniform current density. By taking the curl in cylindrical polars, we found that the vector potential A_z has radial dependence and is given by A_z=-\frac{\mu_0 I}{4 \pi R^2}r^2 + const. We then showed that this satisfies Poisson's equation and also that \nabla \cdot \mathbf{A}=0, corresponding to a specific gauge choice. Lastly, we determined that the total current passing through a cross-section of the wire is equal to \frac{I}{\pi R^2} \mathbf{\hat{z}}.
latentcorpse
A long straight wire of radius R carries a unifrom current density $\mathbf{J}$ inside it.

In the first part of the question I worked out that the magnetic field inside the wire was

$\mathbf{B}=\frac{\mu_0 I r}{2 \pi R^2} \mathbf{\hat{\phi}}$

I am now asked to get the vector potential insde the wire and give the hint of taking the curl in cylindrical polars. So far I have:

$\mathbf{B}=\nabla \wedge \mathbf{A}$
But $\mathbf{A(r)}=\frac{\mu_0}{4 \pi} \int dV' \frac{\mathbf{J(r-r')}}{|\mathbf{r-r'}|}$
so clearly it's parallel to $\mathbf{J}$ which is in the z direction so we conclude that
$A_r=A_{\phi}=0$ and that $A_z$ is non zero.

Taking the curl in cylindrical polars,

$\nabla \wedge \mathbf{A}=\left(\frac{1}{r} \frac{\partial{A_z}}{\partial{\phi}}-\frac{\partial{A_{\phi}}}{\partial{z}} \right) \mathbf{\hat{r}} + \left(\frac{\partial{A_r}}{\partial{z}}-\frac{\partial{A_z}}{\partial{r}} \right) \mathbf{\hat{\phi}} + \frac{1}{r} \left(\frac{\partial}{\partial{r}}(rA_{\phi})-\frac{\partial{A_r}}{\partial{\phi}} \right) \mathbf{\hat{z}}$
which whittles down to:
$\nabla \wedge \mathbf{A}=\frac{1}{r} \frac{\partial{A_z}}{\partial{\phi}} \mathbf{\hat{r}} - \frac{\partial{A_z}}{\partial{r}} \mathbf{\hat{\phi}}$

We have that this must be equal to $\mathbf{B}$ which is given above, so by comparing terms we get

$-\frac{\partial{A_z}}{\partial{r}}=\frac{\mu_0 I r}{2 \pi R^2} \Rightarrow -\int dA_z=\frac{\mu_0 I}{2 \pi R^2} \int r dr \Rightarrow A_z=-\frac{\mu_0 I}{4 \pi R^2}r^2 + const$
and the otehr term gives:
$\frac{\partial{A_z}}{\partial{\phi}}=0 \Rightarrow A_z$ is a constant, but in the line above, we found it to have radial dependence - something isn't quite right here, can anybody help me?

latentcorpse said:
We have that this must be equal to $\mathbf{B}$ which is given above, so by comparing terms we get

$-\frac{\partial{A_z}}{\partial{r}}=\frac{\mu_0 I r}{2 \pi R^2} \Rightarrow -\int dA_z=\frac{\mu_0 I}{2 \pi R^2} \int r dr \Rightarrow A_z=-\frac{\mu_0 I}{4 \pi R^2}r^2 + const$

The 'const' in this equation doesn't have to be a constant in all variables, it just can't have any $r$ dependence (otherwise the partial derivative w.r.t. $r$ of that "constant" term would be non-zero) It can however depend on $\phi$ and $z$ (For example, $\frac{\partial}{\partial r} 5z^2\cos\phi=0$ )

$$\implies A_z=-\frac{\mu_0 I}{4 \pi R^2}r^2+ g(\phi,z)$$

Where $g$ is some unknown function of $\phi$ and $z$
and the otehr term gives:
$\frac{\partial{A_z}}{\partial{\phi}}=0 \Rightarrow A_z$ is a constant, but in the line above, we found it to have radial dependence - something isn't quite right here, can anybody help me?

Again, your constant need not be a constant, it just can't depend on $\phi$

$$\implies A_z=h(r,z)$$

Put the two conditions together and you find that $$A_z=-\frac{\mu_0 I}{4 \pi R^2}r^2 +f(z)$$ satisfies both equations simultaneously for any function $f(z)$...You are free to choose any $f(z)$ you like; as is typical since the vector potential is defined by a first order differential (the curl in this case) and so is only unique up to a 'constant' (in this case, your constant can depend on $z$ without affecting B)

ok. thanks.

next I'm asked to show explicitly that this satisfies Poisson's equation, i.e.

$\nabla^2 \mathbf{A}=-\mu_0 \mathbf{J}$

and also that $\nabla \cdot \mathbf{A}=0$.

For the first part I have:
$\nabla^2 \mathbf{A}=\nabla(\nabla \cdot \mathbf{A})-\nabla \wedge \nabla \wedge \mathbf{A}$ (taking the Laplacian of a vector.

then it makes more sense to show the divergence is 0 now to eliminate that term from the expansion of the laplacian.

$\nabla \cdot \mathbf{A}=\frac{\partial{A_z}}{\partial{z}}=\frac{\partial{f(z)}}{\partial{z}} \neq 0 \forall f(z) \neq 0$
bit confused here?

however assuming that works, we get $\nabla^2 \mathbf{A}=-\nabla \wedge (\nabla \wedge \mathbf{A})=-\nabla \wedge \mathbf{B}$

$\nabla \wedge \mathbf{B}=-\frac{\partial{B_{\phi}}}{\partial{z}}\mathbf{\hat{r}} + \frac{1}{r} \frac{\partial}{\partial{r}} (r B_{\phi}) \mathbf{\hat{z}}$ as $B_r=B_z=0$

now $\frac{\partial{B_{\phi}}}{\partial{z}}=0$ as $\mathbf{B}=\frac{\mu_0 I r}{2 \pi R^2} \mathbf{\hat{\phi}}$

so $\nabla \wedge \mathbf{B}=\frac{1}{r} \frac{\partial}{\partial{r}}(\frac{\mu_0 I r^2}{2 \pi R^2}) \mathbf{\hat{z}}=\frac{\mu_0 I}{\pi R^2}\mathbf{\hat{z}}$

$\Rightarrow\nabla^2 \mathbf{A}=-\mu_0 \frac{I}{\pi R^2}\mathbf{\hat{z}}$

now my trouble is explaining why $\mathbf{J}=\frac{I}{\pi R} \mathbf{\hat{z}}$. Clearly the direction is fine and the untis seem ok as we have amperes/metre^2 - but there must be some definition I'm meant to use that I'm just missing in my notes?

Last edited:
latentcorpse said:
$\nabla \cdot \mathbf{A}=\frac{\partial{A_z}}{\partial{z}}=\frac{\partial{f(z)}}{\partial{z}} \neq 0 \forall f(z) \neq 0$
bit confused here?

Requiring $$\vec{\nabla}\cdot\vec{A}=0$$

corresponds to a specific gauge choice (it removes some of the freedom you normally have in chossing your 'constants'); so if you want to choose $$\vec{A}$$ in a way that makes this true, just choose an $f(z)$ with zero divergence (the simplest choice is just $f(z)=0$)

now my trouble is explaining why $\mathbf{J}=\frac{I}{\pi R^2} \mathbf{\hat{z}}$. Clearly the direction is fine and the untis seem ok as we have amperes/metre^2 - but there must be some definition I'm meant to use that I'm just missing in my notes?

Well, what is the total current passing through a cross-section of the wire if the volume current $$\vec{J}$$ is uniform and runs in the z-direction?...since the total current is required to be $I$ equate the two expressions and solve for $J$.

$I=\int_S \mathbf{J} \cdot \mathbf{dA} \Rightarrow I=|\mathbf{J}| \pi R^2 \Rightarrow |\mathbf{J}|=\frac{I}{\pi R^2}$

cheers m8.

## 1. What is the magnetic potential of a long straight wire?

The magnetic potential of a long straight wire is the measure of the energy required to move a unit magnetic pole from infinity to a point along the wire.

## 2. How is the magnetic potential of a long straight wire calculated?

The magnetic potential of a long straight wire can be calculated using the formula: μ0I/2πr, where μ0 is the permeability of free space, I is the current flowing through the wire, and r is the distance from the wire.

## 3. What factors affect the magnetic potential of a long straight wire?

The magnetic potential of a long straight wire is affected by the strength of the current, the distance from the wire, and the permeability of the surrounding medium.

## 4. How does the direction of the current affect the magnetic potential of a long straight wire?

The direction of the current affects the direction of the magnetic field, which in turn affects the magnetic potential. Reversing the direction of the current will also reverse the direction of the magnetic potential.

## 5. What are some real-world applications of the magnetic potential of long straight wires?

The magnetic potential of long straight wires is utilized in various technologies such as electric motors, generators, and speakers. It is also used in medical imaging, such as MRI machines, which rely on the magnetic potential of wires to create images of the body's internal structures.

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