Amplitude of antenna-voltages induced by microwaves
It's not too complicated to calculate.

Imagine you have some kind of transmitter that transmits power equally in all directions. This is called an isotropic antenna. Such a thing can't exist, for a radio transmitter, but for the moment let's assume it does.

If this transmitter has a power of 1W, then for any sphere centered on that transmitter (assuming the sphere is far enough away to be in the far field), there is 1W passing through that sphere, too. The sphere must intercept all the transmitted power because there is no direction from the transmitter that doesn't hit the sphere.

Now imagine that the distance between the receiver and the transmitter is the same as the radius of the sphere.

The surface area of a sphere is given by:

$$ A = 4\pi r^2 $$

So if the receiver and transmitter are 100 meters apart, then imagine a sphere with a radius of 100 meters, and a surface area of

$$ A = 4 \pi 100^2 = 125664\:\mathrm{m^2} $$

If the transmitter were transmitting with a power of 1W, then at the receiver's distance, that 1W of power is spread over the area of the sphere. If we divide the transmitter's power by the sphere's area, we get the irradiance:

$$ { 1\:\mathrm{W} \over 125664\:\mathrm{m^2} } = 7.96\:\mathrm{\mu W / m^2} $$

This means if you had a "net" for electromagnetic energy that was 1 square meter large, and it was 100 meters away from a 1 watt transmitter that radiated energy equally in all directions, that net would intercept 7.96 microwatts of power.

But you don't have a net, you have an antenna. How big of a net is formed by an antenna? The answer to this question is the effective aperture, which is actually another way of specifying the gain (G) of the antenna:

$$ A_e = { \lambda^2 \over 4\pi } G $$

λ is the wavelength of the radiation in question.

The transmitting antenna also has some gain (remember how an isotropic antenna can't exist?). If the transmitting antenna has a gain of 3 dBi, that means a doubling of power relative to an isotropic radiator. So the irradiance will be twice what was calculated before, or 15.9 microwatts per square meter.

So far we've calculated the irradiance at the receiver, and how much power will be received by that receiver, given its effective aperture. But you didn't ask about power, you asked about voltage.

Electrical power (P) is the product of voltage (E) and current (I):

$$ P = IE $$

Also, in free space, the ratio of voltage and current is fixed by the impedance of free space, which is about 377 ohms, which means (by Ohm's law):

$$ {E \over I} = 377\:\Omega $$

We know the irradiance (15.9 microwatts per square meter), so we can solve this system of equations to get:

$$ \begin{align}
P &amp;= 15.9\:\mathrm{\mu W / m^2} \\
I &amp;= 205\:\mathrm{\mu A / m} \\
E &amp;= 77\:\mathrm{m V / m} \end{align} $$

That value for E (77 millivolts per meter) is almost what you asked for. It is the potential gradient, meaning if you measured the rate of change in voltage per unit distance at one point, it would be 77 millivolts per meter.

But you didn't ask about the potential gradient. You asked about the voltages in a piece of metal.

Here's where it gets very complicated (I lied about it being simple). It will still be true that the antenna will intercept some fraction of the transmitted power. We can calculate the power intercepted by calculating the irradiance at the receiving antenna, and multiplying that by the effective aperture of the receiving antenna. And the ratio of voltage to current is the impedance of the antenna, whatever that might be. It probably won't be the 377 Ω it is for free space.

For simple antenna designs, we do know the impedance. For example, we know that the impedance of a half-wave dipole at the feedpoint is around 73 Ω. So we could do nearly the same calculations as before, but when we include the effective aperture there's an extra meter unit (remember effective aperture is measured in square meters), and that will cancel the "per meter" part we got when we calculated the voltage gradient. You'll be left with just a voltage in volts, instead of a voltage gradient in volts per meter.

That tells you the voltage at the feedpoint. You can also calculate the voltage at other parts of the dipole, if you know the impedance at that point. A dipole's impedance gets higher the closer you get to the tips. The power is the same, but if the impedance goes up that must mean the voltage goes up and the current goes down.

But what if you don't have a dipole, but instead some random piece of metal?

I'm afraid there's no easy answer at this point. By calculating the irradiance at the metal, you'll have some idea of how much power the metal might intercept. I say "some idea" because the effective aperture of an antenna isn't necessarily the physical size of the antenna. A dipole for example is a very thin wire, but it intercepts power from a larger area around the wire.

And we won't know what the impedances are at various points on the metal. This is going to depend on the geometry of the metal, among other things.

If you want the exact answer, you'll need to solve Maxwell's equations, which isn't so easy. Or use a computer to simulate the equations for you.