# Inductance and Capacitance of wire near ground plane

I'm stuck on the following question and would greatly appreciate any help:

A lossless transmission line is formed by a wire of radius a placed in vacuum a distance d above an infinite conducting plane with $d>>a$. Calculate the capacitance and Inductance per unit length.

I'm able to calculate the capacitance per unit length. I let the wire at z = d carry a charge +Q uniformly distributed across its length l (I know this is in general not true but for $d>>a$ the uniform distribution should be a good approximation). As the plate is conducting I then calculate the electric field above the plane by placing an image wire carrying charge -Q at z = -d on the other side of the plane. This allowed me (after fiddling around with the signs to get the correct answer) to integrate the electric field from $0$ to $d-a$ to get the following potential between the plane and the wire:

$$V=\frac{Q}{2\pi\epsilon_0l}\ln\frac{2d}{a}$$

so therefore:

$$C' = \frac{2\pi\epsilon_0}{\ln\frac{2d}{a}}$$

However the Inductance per unit length is what I'm stuck on. At first I tried to calculate the magnetic field of the wire:

$$B = \frac{\mu_0I}{2\pi r}$$

Then I calculated the magnetic field of a large plane of width w carrying a current I:

$$B_{sheet}=\frac{I}{w}$$

so because the conducting plane is described as infinite in the question I took $B_{sheet} = 0$ as $w\rightarrow \infty$. Then I calculated the flux between the wire and the plane:

$$\phi = l\int_{a}^{d}\frac{\mu_0I}{2\pi r}=\frac{\mu_0 I l}{2\pi}\ln\frac{d}{a}$$

However this is not the right answer as I seem to be missing a factor of 2 inside the $ln$.

How do I approach the calculation of the Inductance per unit length to get the correct answer?

• Some information here? zakii.la.coocan.jp/tline_e/14_microstripline_z0.htm – Farcher May 8 '18 at 10:18
• Thanks very much for the info. This raises another question though: Why is it possible to apply the image technique to a current distribution near a conductor? As far as I understand it the image charge method works because the potential is $0$ on the surface and inside a conductor. However I don't think there should be any restriction on the magnetic field, unlike the electric field, near or inside the conductor. – 1MegaMan1 May 8 '18 at 12:13
• There can be no magnetic field inside a perfect conductor so any magnetic field must be parallel to the surface. A “negative” current carrying image conductor the same distance behind the plane conducting surface achieves this, – Farcher May 8 '18 at 12:45
• Sorry if I’m missing something very obvious, but why can there be no magnetic field inside a perfect conductor? I know the reasoning for the absence for an electric field, but can’t quite figure out the reasoning for the case of a magnetic field. – 1MegaMan1 May 8 '18 at 14:14