# Where does the excess energy emitted by a microwave go?

If there is nothing in the microwave, where does the excess radiation go? Why doesn't the radiation accumulate and blow it up?

Should I cook two pieces of Canadian Bacon twice as long as I cook one?

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bacon should be covered in another question. –  troyaner Aug 15 '11 at 18:15

The power going into an empty microwave will accumulate as an EM field inside the cavity up to the point where the input power equals the rate of leakage. Energy inside the cavity will leak out through the little holes in the metal screen in the door, through ohmic losses in the sided of the microwave, feedback into the power source, etc. The rate of power leakage from the cavity is given by $P_{leak} = 2\pi E f / Q$ where $E$ is the energy stored in the cavity, $f$ is the resonant frequency of the cavity and Q is the "quality factor". In equilibrium then, $E = QP_{input}/{2\pi f}$. For an resonant frequency of around 2.5 GHz, input power of 1kW, Q of say 100 (guess), the accumulated energy is only about a milli-joule. Of course, the materials making up the microwave probably aren't designed to dissipate 1kW of power over an extended time, so eventually the microwave may explode or something due to overheating of components.

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I seem to recall hearing that the power would mainly return to the magnetron and cause internal arcing. –  Kevin Reid Aug 16 '11 at 0:22
Thanks for a good explanation, but... Can you explain a bit where this equation came from? Like why would it radiate more with a higher resonant frequency? Did you find it somewhere? Aren't the bars on the front spaced to block a specific wavelength of light? –  user420667 Aug 18 '11 at 2:50
See the wikipedia page on the quality factor: en.wikipedia.org/wiki/Q_factor . The mesh in front is designed to block the microwaves while letting visible light through, but it won't be 100% efficient. –  user1631 Aug 18 '11 at 18:03