Factor of two for inductance of parallel plates Studying transmission lines I am confused about a "missing" factor of two in the formula $L = \mu_0 a/b$ for the inductance per unit length of two "infinite" parallel sheets.
Say we have two parallel conducting plates with equal and opposite currents $I$. Let the width of the plates be $b$, the distance between them $a$ and length $l$. Assume $a \ll b \ll l$, so that the field inside is uniform. Placing an Amperian loop around the whole arrangement, the net current is zero, so there should be no external field in our approximations. Having a loop around one of the plates,
$$ \int \vec{B} \cdot d\vec{l} = \mu_0 I \rightarrow B = \frac{\mu_0 I}{ b} $$
This is the field from one of the plates, so the total field is
$$\frac{2\mu_0 I}{b}$$ 
giving inductance
$$ L = \frac{\Phi_m}{I} = \frac{2\mu_0 \Phi a}{b} =2 \frac{\mu_0 a}{b} $$
per unit length. This is a factor of $2$ off from the true value. Can somebody help resolve my confusion? 
 A: Let $B_0 = \mu_0 I / b$. Then Ampere's law only says that for each plate,
$$B_{\text{right}} - B_{\text{left}} = B_0$$
where $B_{\text{right}}$ and $B_{\text{left}}$ are the fields on the left and right side of the plate. In the case of a single plate, assuming standard boundary conditions, the fields are
$$-B_0 / 2 \quad |\text{plate}| \quad B_0/2.$$
In this case with two plates, with the same boundary conditions, the fields are
$$0 \quad |\text{plate 1}| \quad B_0 \quad |\text{plate 2}| \quad 0.$$
The fields of the two plates add up inside, $B_0 / 2 + B_0 / 2 = B_0$, and cancel out outside. There is no factor of $2$.

The reason this is confusing is that for a single plate, there are different boundary conditions. For example, the configuration
$$0 \quad |\text{plate 1}| \quad B_0$$
for the left plate indeed satisfies Ampere's law; I think this is what you were implicitly doing when you considered one plate in isolation. In fact, this is exactly the solution you would want if, say, there was a superconductor to the left of the plate. Similarly for the other plate you might take
$$B_0 \quad |\text{plate 2}| \quad 0.$$
Then the total field configuration looks like
$$B_0 \quad |\text{plate 1}| \quad 2 B_0 \quad |\text{plate 2}| \quad B_0.$$
But typically when we solve these problems we're in a situation where the magnetic field is zero at infinity. So while this configuration satisfies Ampere's law it's not the one we're looking for, it's what we would get if there were an additional constant background field. Incidentally, you can still use this solution to calculate the inductance as long as you remember that the $\Phi$ in the inductance formula is the change of the flux when you turn the current on. Without the current the field is $B_0$ and it increases to $2 B_0$. Since $2 B_0 - B_0 = B_0$ there's again no factor of $2$.
