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I am reading JB Pendry's original paper on metamaterials (Magnetism from Conductors, and Enhanced Non-Linear Phenomena) and I am really having some trouble understanding just a few of his mathematical steps and was wondering if anyone could help me out here.

So he begins by saying that the field inside a cylinder in a unit cell is given by

\begin{equation} H = H_0 + j -\frac{\pi r^2}{a^2}j \end{equation} where $a$ is the dimension of the unit cell, $j$ is the induced current per unit length caused by the external field, $r$ is the radius of the cylinder and $H_0$ is the external field applied in a direction that is axial to the cylinder. He then says that the emf around the circumference of the cylinder is: \begin{equation} emf = -\pi r^2 \mu_0 \frac{\partial}{\partial t}\left[ H_0 + j - \frac{\pi r^2}{a^2} j \right] - 2\pi r \sigma j \end{equation} where $\sigma$, he says, is the resistance of the cylinder surface per unit area.

I have issue with both of these terms. The first term obviously comes from faradays law, but why on earth is it multiplied by the area of the cylinder cap?? EDIT: Figured this one out, super simple. I forgot the form of the induced emf is a function of the flux, which is an integral over the area. Doy!

The second term is the voltage drop due to the resistance of the cylinder. But he said $\sigma$ is the resistance per unit area, but then proceeds to multiply by only a single dimension unit, namely $r$. So at the end of the day, the units don't add up to be volts.

What's going on here???

DK2AX
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user41178
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1 Answers1

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From interface conditions on the cylinder boundary we get $H_{inside}^z-H_{outside}^z=j^{\varphi}$.

Using equation for flux, $\phi=\int\int_S B\cdot dS$, we get $$\frac{1}{a^2\mu_0}\phi=\frac{r^2\pi}{a^2}H^z_{inside}+ \left(1-\frac{r^2\pi}{a^2}\right)H^z_{outside}=H_0^z$$ because field lines of induced fields are closed, so they do not contribute to the total flux, only exerted field, $\vec{H}_0=H^z_0\hat{z}$, produce net flux. From previous two equations, using elementary algebra, we get: $$H^z_{inside}=H_0^z+j^{\varphi}-\frac{r^2\pi}{a^2}j^{\varphi},$$ and $$H^z_{outside}=H_0^z-\frac{r^2\pi}{a^2}j^{\varphi}.$$ As you realised, units are not balanced in equation for $\textit{emf}$ and I don't know how to fix that, except that $\sigma$ is pure resistance and not resistance per unit area as is stated in the paper.

dbojanjac
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