# Calculation of inductance of rectangle coil

I'm currently working on a project about inductance. I want to calculate the inductance of a rectangle-shaped coil with a pitch different from zero. I tried this by applying the law of Biot-Savart. If i do this for one wire in a cartesians axis system i get a formula of the shape $B=\int{\frac{(y-y_0)}{\sqrt{((x-x_0)^2+(y-y_0)^2)}^3}}$ from 0 to $x_0$ or $y_0$. Which i can't integrate by hand or by numerical integration in matlab (singularity at $y_0$). If i write it out in polar coordinate system, i get $B=constant*(\sin(\theta_2)-\sin(\theta_1))$. But i can't figure out how to calculate the magnetic flux (phi=int{B dA}). And numerical integration has again some problems with the singularity close to the wire.

Has someone a better idea to calculate the inductance instead of using Biot-Savart? Or someone who knows how to integrate those integrands. Thanks a lot!

• I added some proper math markup. Let me know if there are any issues. We use mathjax, which supports normal LaTeX commands, so you can probably fix it yourself. – Connor Wolf Aug 28 '14 at 1:49
• I need some more info....1) Is that cube under the radical? 2)Is this a double integral from 0 to x0 and 0 to y0? 3)Can you give additional details on what you mean by pitch?(Can you provide a picture) – tman Aug 28 '14 at 14:06
• 1) i'm not sure what u mean, but my integration area is under the line. 2) It is a double integral 3) if you have multiple windings, the pitch is the distance between the centers of two adjacent windings which is different from the diameter due to isolation of the wire – user3329103 Aug 28 '14 at 14:32

There is a web-calculator for this. And here, you can find an excellent paper about methods of calculation for different shapes of coils.

Consider a line in length of $\dfrac{y_0}{2}$ carrying DC current $I$. We wish to find the magnetic field at the point $P\left(\dfrac{x_0}{2},0,0\right)$.

The Biot Savart formula:

$$\mathbf{B} = \dfrac{\mu I}{4 \pi} \int \limits_C \dfrac{d\mathbf{\ell} \times \mathbf{a_R}}{R^2} = \dfrac{\mu I}{4 \pi} \int \limits_C \dfrac{d\mathbf{\ell} \times \mathbf{R}}{R^3}$$

Substitute the quantities:

$$\mathbf{\ell} = y \; \mathbf{a_y}, \quad \mathbf{R} = \dfrac{x_0}{2}\mathbf{a_x} - y \; \mathbf{a_y}, \quad d\mathbf{\ell} \times \mathbf{R} = \dfrac{x_0}{2}(-\mathbf{a_z})$$

Considering that the path $C$ is from the bottom to top of the wire, it becomes:

$$\mathbf{B} = \dfrac{\mu I}{4 \pi} \int \limits_C \dfrac{d\mathbf{\ell} \times \mathbf{R}}{R^3} = \dfrac{\mu I}{4 \pi} \int \limits_C \dfrac{\dfrac{x_0}{2}(-\mathbf{a_z})}{R^3} = -\dfrac{\mu I}{8 \pi} \int \limits_C \dfrac{x_0\mathbf{a_z}}{R^3} = -\dfrac{2 \mu I y_0}{\pi x_0 \sqrt{x_0^2 + y_0^2}} \mathbf{a_z}$$

If the entire structure is a rectangular of $x_0$ by $y_0$, just sum all four line segments due to the superposition theorem:

$$\mathbf{B} = -2\dfrac{2 \mu I y_0}{\pi x_0 \sqrt{x_0^2 + y_0^2}} \mathbf{a_z} - 2\dfrac{2 \mu I x_0}{\pi y_0 \sqrt{x_0^2 + y_0^2}} \mathbf{a_z} = -\dfrac{4 \mu I}{\pi x_0 y_0 \sqrt{x_0^2 + y_0^2}} \left[ x_0^2 + y_0^2 \right] \mathbf{a_z} = \dfrac{4 \mu I}{\pi x_0 y_0 \sqrt{x_0^2 + y_0^2}} \left[ x_0^2 + y_0^2 \right] (-\mathbf{a_z}) = \dfrac{4 \mu I}{\pi x_0 y_0 \sqrt{x_0^2 + y_0^2}} \left[ x_0^2 + y_0^2 \right] \mathbf{a_{-z}}$$