Fall 2007 14 Em

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 10oC, 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 $\epsilon \approx 80+10i$.

***

Answer

***

(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)
\begin{align} P=\frac{1}{2}\sigma E^2 = \frac{1}{4}f\epsilon_2E^2. \end{align}

Therefore,

(2)
\begin{align} E=2\sqrt{P/f\epsilon_2} \end{align}

(b) Substituting the numbers, we get

(3)
\begin{align} E[\mbox{statV/cm}]\approx 2 \sqrt{10^7[\mbox{erg/cm^3}]/2.45 \times 10^9[\mbox{Hz}]/10} = 0.04 [\mbox{statV/cm}]=1.2\mbox{kV/m} \end{align}

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 Eout in the rest of the oven is larger:

(4)
\begin{align} E_{out}=E[1+(\epsilon-1)N] \end{align}

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

(5)
\begin{align} |Eout| \sim |\epsilon E|/3 \approx 30|E|. \end{align}

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

Add a New Comment
or Sign in as Wikidot user
(will not be published)
- +
Unless otherwise stated, the content of this page is licensed under Creative Commons Attribution-Share Alike 2.5 License.