Biot-Savart law and Poisson's equation


by Mbert
Tags: biotsavart, equation, poisson
Mbert
Mbert is offline
#1
Aug18-11, 07:35 PM
P: 64
Dear colleagues,

I have questions regarding Biot-Savart law. From [1], it is shown that the equation (Biot-Savart) is derived from the solution to Poisson's equation (assuming here div A=0)
\begin{equation}
\vec{\nabla}^2 \vec{A} = -\mu \vec{J}
\end{equation}
which is
\begin{equation}
\vec{A}(\vec{r}) =\frac{\mu}{4\pi}\int_V{\frac{\vec{J}{\rm d}^3r'}{\left|\vec{r}-\vec{r'}\right|}}
\end{equation}

where $\vec{r}$ is the position where $\vec{A}$ is evaluated and $\vec{r'}$ is the position where the integral is evaluated.

The first thing that troubles me is the singularity $\left|\vec{r}-\vec{r'}\right|$ when we evaluate the field at the point of integration. For a wire of finite radius, this means that the $\vec{A}$ field inside the conductor is infinite (or am I missing something?). If so, why in books on electromagnetics do we usually replace the conductor by an equivalent filamentary current $I=\vec{J}\cdot{\rm d}\vec{s}$? The field calculated inside the conductor will be different. This can be seen from the solution for the $\vec{B}$ field by using Ampere's equation (for the infinitely-long finite-radius wire)
\begin{equation}
B_{\theta}=\frac{\mu I}{2 \pi \rho}
\end{equation}outside the wire
\begin{equation}
B_{\theta}=\frac{\mu I \rho}{2 \pi R_{wire}^2}
\end{equation} inside the wire

where $R_{wire}$ is the cross-section radius and $(\rho,\theta,z)$ are the cylindrical coordinates. This means essentially that the field at $\rho=0$ is zero and that it is proportional to $\rho$ inside the conductor and inversely proportional to $\rho$ on the outside. How can we get this from the solution to Poisson's equation for a finite-radius wire?

The other thing that troubles me with the solution to Poisson's equation (second equation) is the value of the integrand when $\vec{r}=\vec{r'}$, but outside the wire (thus where J=0). This means we get a 0/0 integrand for each $\vec{r}$ outside the wire, which numerically gives NaN for the whole integral. Is this a problem analytically? because this contribution might (should) be 0, probably by using L'Hopital's rule (I guess).

Best regards,
M.

[1] Smythe,W.R., "Static and dynamic electricity", McGraw-Hill, 1968.
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Bill_K
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#2
Aug20-11, 01:53 PM
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M, I know the Poisson integral looks singular, but it is not. What you're forgetting about is the volume element. To see the behavior near r = r', write the integrand near that point in terms of a spherical coordinate, R = |r - r'|. Then d3r = 4π R2 dR, and the integral is like ∫ 4π J(0) R dR, which is nonsingular. The factor in the numerator that comes from the volume element goes to zero faster than the denominator does.
Mbert
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#3
Aug20-11, 04:24 PM
P: 64
How did the [itex]{\rm d}\theta[/itex] and [itex]{\rm d}\phi[/itex] disappear in spherical coordinates? Isn't that equivalent of considering an infinitesimal rectangular prism [itex]xy{\rm d}z[/itex] (which would also be an infinitesimal volume, but not infinitesimal in all three dimensions)?

M.

Bill_K
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#4
Aug21-11, 04:41 AM
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Biot-Savart law and Poisson's equation


In the immediate neighborhood of R = 0, the integrand is spherically symmetric and you can integrate at once over solid angle, producing the factor of 4π. In other words, ∫∫∫ ... d3r = ∫∫∫ ... R2 dR d2Ω = ∫ ... 4π R2 dR
Mbert
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#5
Aug21-11, 08:48 AM
P: 64
As posted previously in the forums here:

http://www.physicsforums.com/showthread.php?t=119419

the field of an infinite wire at r=0 is infinity. In your case, are saying that this field should be zero?

From Green's function, it is normal to get infinite vector potential, since we assumed Dirac sources. So the filamentary current case is ok with me. However, what happens inside with a finite conductor (e.g. a cylindrical conductor)? Wouldn't that mean we get infinity everywhere inside? Or this is perhaps where I mix things up.

thanks

M.
vanhees71
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#6
Aug21-11, 12:27 PM
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According to [itex]\vec{j}=\sigma \vec{E}[/itex] you get a finite electric field inside a finite conductor.

The solution of this standard magnetostatics problem, using the above (approximate, i.e., non-relativistic form of) Ohm's Law, for an infinite wire is a constant electric field along the wire.
Mbert
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#7
Aug21-11, 02:50 PM
P: 64
I'm interested in the magnetic field B, not the electric field E. That's why I'm interested in the vector potential A, found from the solution of Poisson's equation.

M.


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