### Fall 2007

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A small amount of water is being warmed in a microwave oven.

(a) Derive the formula for the amplitude *E* of the electric field in terms of the power *P* dissipated in a unit volume of water, microwave frequency *f* (in Hz), and the complex dielectric function of water $\epsilon=\epsilon_1+i\epsilon_2$. Assume that the field penetrates the water uniformly.

(b) Compute the voltage drop *V* (assuming uniform field) across the longest dimension of the oven, *L* = 0.3 m. Use the following information. Typically, it takes about a minute to heat a cup of water by 10^{o}C, so that $P \approx 10^6 W/m^3$. The microwave frequency is $\omega=2\pif$, where *f* = 2.45GHz. At such frequency the dielectric function of water is $\epsilon \approx 80+10i$.

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Answer

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(a) Let $\sigma$ be the real part of the conductivity. In the Gaussian units $\sigma=\omega \epsilon_2/4\pi=f\epsilon_2/2$. The ac Joule heating is

(1)Therefore,

(2)(b) Substituting the numbers, we get

(3)Accordingly, the voltage drop across the oven is $V=EL\approx 360 \mbox{V}$.

One more consideration is in order to get the correct estimate for *V*. (However, failure to acknowledge it was not penalized by taking off points in grading the exam).

Strictly speaking, the field *E* we computed is actually the field inside the water. The magnitude of the field *E _{out}* in the rest of the oven is larger:

where *N* is referred to as the depolarization factor. If water forms a very shallow puddle and the microwave field is parallel to its surface, i.e., horizontal, *N* is very small and $E_{out}\sim E$, so the above estimate of *V* stands.

However, if the dimensions of the volume occupied by the water are comparable (e.g, water filling a common mug), then *f* ~ 1/3, so that

Therefore, the actual total voltage across the microwave oven is ~ 360V × 30 ~ 10 kV. This explains why safety features are necessary.

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