# Mutual inductance using bessel function

1. Mar 19, 2009

### salla2

Hi, I am en electrical engineering grad student and I have to solve an equation to calculate the mutual inductance between an antenna and a micro-inductor. I think it is a Bessel equations but I don't know how to solve.

M(a,b,d)=(1.45x10^-8)*integral(J1(1.36x)*J1(0.735x)*exp(-x-13.6))dx
the integration is from 0 to infinity

2. Mar 21, 2009

### coomast

Hello salla2,

I have rewritten your problem as:

$$I=d e^{-c} \cdot \int_{0}^{\infty}e^{-x}J_1(ax)J_1(bx)dx$$

Leaving out the factor before the integral it comes down to:

$$I=\int_{0}^{\infty}e^{-x}J_1(ax)J_1(bx)dx$$

This is not easy. I tried several ways.

*) The online mathematica integrator: NOK
*) xmaxima: NOK
*) The following integral book:
"Integraltafel, zweiter teil bestimmte integrale" by grobner - hofreiter
gives on page 203 the following formula:

$$\int_{0}^{\infty}e^{-ax}J_{v}(bx)J_{v}(cx)dx=\frac{1}{\pi\sqrt{bc}}Q_{v-\frac{1}{2}} \left( \frac{a^2+b^2+c^2}{2bc}\right)$$
$$Q_{v-\frac{1}{2}}(z)=\frac{\sqrt{2\pi}\Gamma\left(v+\frac{1}{2}\right)} {2^{v+1}\Gamma\left(v+1\right)} \cdot z^{-v-\frac{1}{2}} F\left(\frac{2v+1}{4},\frac{2v+3}{4},v+1;\frac{1}{z^2}\right)$$
Which is a "Legendresche Function" F is the hypergeometric series.

*) The following book:
"Table of Laplace Transforms" by Roberts and Kaufman
gives on page 57, the following function for the Laplace transform of $J_{v}(at)J_{v}(bt)$
which can be used considering the definition of the Laplace transform, s is then afterwards set to 1.

$$\frac{1}{\pi\sqrt{ab}}Q_{v-\frac{1}{2}} \left( \frac{s^2+a^2+b^2}{2ab}\right)$$
The definition of $Q_{v-\frac{1}{2}}$ was given as:
$$Q_v^{\mu}(x)=\frac{e^{\mu \pi i}\pi^{1/2}\Gamma\left(\mu+v+1\right)(x^2-1)^{\mu/2}} {2^{v+1}\Gamma\left(v+\frac{3}{2}\right)x^{\mu +v+1}} \cdot {\left.}_2F_1{\right.} \left[\frac{\mu+v+1}{2},\frac{\mu+v+2}{2};v+\frac{3}{2};\frac{1}{x^2}\right]$$
In which $\mu=0$

I can't help you any further on this, hope it helps a bit. It is a difficult question...

coomast

3. Mar 24, 2009

### salla2

I didn't understand a single word but thanks for your help anyway!!