My goal is to create an expression that gives the field strength of an arced segment of wire at a fixed distance away from the center point of the arc. To be used in EMF (electromotive force) visualization. The following Equation gives this relation:
\begin{equation} d\beta=\frac{\mu_{0}I}{4\pi}\int_{_{C}}^{}\frac{\overrightarrow{dl}\times \overrightarrow{r'}}{\left| r' \right|^{3}} \end{equation}
Source: Biot-Savart Law
However, I have been having issues getting this into an algebraic expression that can be easily implemented into Code. as when I
represent the cross products of dl and r’ as an expression over the magnitude of r’^3 The simplest I've been able to get it is as follows:
\begin{equation}
d\beta=\frac{\mu_{0}I}{4\pi}\int_{\alpha_{min}}^{\alpha_{max}}\frac{(cos(\alpha) z)\widehat{i}+(sin(\alpha)z)\widehat{j}+(cos(\alpha)x+sin(\alpha)y+r)\widehat{k}}{(x^{2}+y^{2}+z^{2}+r^{2}+2r(sin(\alpha)y+cos(\alpha)y)^{\frac{3}{2}})}d\alpha
\end{equation}
Steps are shown in the image link below.
Which makes the integration stupidly long. As A and b are equivalent I am trying to use formula a as opposed to b so I don’t end up with a to the power of 3/2 to make integration easier. I’ve been staring at this problem for about 3 weeks now and have tried numerous different thing representing it with respect to beta instead of alpha. As shown in the image below:
The simplest I’ve got it is with respect to alpha as shown on the right-hand side of the page. But the integration is stupidly long. Does anyone have any ideas as to how to get that integral in simplest forms? Ideas I’ve had are switching coordinate systems and right now I somehow ended up in a mixture of cylindrical and Cartesian coordinates.
Please let me know if any additional information is needed.
Best Answer
$\newcommand{\bvec}[1]{\mathbf{#1}}$ $\newcommand{\uv}[1]{\bvec{\hat{e}}_{#1}}$ Work in the traditional Heaviside–Gibbs vector calculus with cylindrical coordinates $(\rho,\phi,z)$. Let $\bvec{J}$ denote the vector-valued distribution $$\bvec{J}(\rho,\phi,z)=I \delta(\rho-a)\delta(z)[\phi_1<\phi<\phi_2]\uv{\phi}\text{.}$$ Then the unique suitably regular vector field $\bvec{B}$ satisfying $$\begin{align} \bvec{\nabla}\cdot \bvec{B} &= 0 & \bvec{\nabla}\times \bvec{B} &= \mu_0 \bvec{J} \end{align}$$ is
$$\bvec{B}(\rho,\phi,z) =\frac{\mu_0 I}{4\pi\sqrt{(\rho+a)^2+z^2}}\left.\left( b_{\rho}(\rho,\phi,z;\phi')\uv{\rho} +b_{\phi}(\rho,\phi,z;\phi')\uv{\phi} +b_z(\rho,\phi,z;\phi')\uv{z} \right)\right\rvert_{\phi'=\phi_1}^{\phi_2}$$
where
$$\begin{align} b_{\rho}(\rho,\phi,z;\phi') &= \frac{z}{\rho}\left(\frac{a^2+\rho^2+z^2}{(\rho+a)^2+z^2}\Pi\left(\tfrac{\phi'-\phi-\pi}{2},-m(\rho,z)\vert m(\rho,z)\right)-F\left(\tfrac{\phi'-\phi-\pi}{2}\vert m(\rho,z)\right)\right)\\ b_{\phi}(\rho,\phi,z;\phi') &=-\frac{z}{\rho}\sqrt{\frac{(\rho+a)^2+z^2}{a^2+\rho^2+z^2-2a\rho\cos(\phi-\phi')}} \\ b_{z}(\rho,\phi,z;\phi') &= \frac{a^2-\rho^2-z^2}{(\rho+a)^2+z^2}\Pi\left(\tfrac{\phi'-\phi-\pi}{2},-m(\rho,z)\vert m(\rho,z)\right)+F\left(\tfrac{\phi'-\phi-\pi}{2}\vert m(\rho,z)\right) \\ m(\rho,z) &= \frac{4a\rho}{(\rho+a)^2+z^2}\text{.} \end{align}$$ Here $F(\phi\vert m)$ and $\Pi(\phi,n\vert m)$ are Legendre's incomplete elliptic integrals of the first and third kind in the amplitude-parameter convention $$\begin{align} F(\phi\mid m)&=\int_0^{\phi}\frac{\mathrm{d}\theta}{\sqrt{1-m\sin^2 \theta}}\\ \Pi(\phi,n\mid m)&=\int_0^{\phi}\frac{\mathrm{d}\theta}{(1-n\sin^2\theta)\sqrt{1-m\sin^2 \theta}}\text{.} \end{align}$$ This solution in terms of elliptic integrals alone is due to Alberto-Gonzalez and Elam-Cardenas (2021), although the basic idea is a few decades older and the case $\phi_2=\phi_1+2\pi$ is classical. You can find the derivation in their paper: it is a few pages long of computer-assisted algebra.
Alberto Gonzalez, Migdonio, and Dorindo Elam Cardenas. 2021. “Analytical Expressions for the Magnetic Field Generated by a Circular Arc Filament Carrying a Direct Current.” IEEE Access 9:7483–95.