Why does an open ended wire cause the load to the oscillator to increase?

Question

Using an ammeter between the wall and signal generator, I noticed that when the lead from a single channel is attached to the signal generator, the power into the signal generator increases by about 100 milliwatts.
When the wire leads are removed from the signal generator, the input power into the signal generator decreases by 100 milliwatts.
Why would an open ended wire cause a load? Is there a way for that wire not to add to the input load?
The lead wire from the signal generator is not attached to anything on the open end.

Kregnach · Answer

There are waves in the wire, and they can be reflected back from the closure.

As an example: when I was working with preamplifiers to increase the signal measured in the oscilloscope, I had to use a 50 ohm closure at the end of every wire, to get a good enough reflection otherwise the signal kind of "leaked" , got noisy.

I guess some electrodynamics calculation could show you what's going on! :D

Bruce-TPU · Answer

It appears that someone who is familiar with RF answered my question elsewhere. His answer was that the attached BnC cable to the signal generator is not an open circuit as I has assumed but because of the Mhz frequency, it actually appears to the signal generator as a mismatched circuit (impedance mismatch) drawing a load.
The signal generator is rated at a 50 ohm impedance and the BnC connector is also 50 ohms. So perhaps it simply sees it as a matched impedance load?

Why does an open ended wire cause the load to the oscillator to increase?

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