Subfield: Electromagnetism

The Physics of a Microwave Oven

This question discusses the generation of microwave radiation in a microwave oven, and its use to heat up food. The microwave radiation is generated in a device called "magnetron". Part A concerns the operation of the magnetron, while part B deals with the absorption of microwave radiation in food.
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Figure 1

$$
\begin{gathered}
d=1 \mathrm{~mm} \\
l=6 \mathrm{~mm} \\
R=7 \mathrm{~mm} \\
h=17.5 \mathrm{~mm}
\end{gathered}
$$

Part A: The structure and operation of a magnetron

A magnetron is a device for the generation of microwave radiation, either in pulses (for radar applications), or continuously (e.g., in a microwave oven). The magnetron has a mode of self-amplifying oscillations. Supplying the magnetron with static (non-alternating) voltage quickly excites this mode. The microwave radiation thus created is transmitted out of the magnetron.

A typical microwave oven magnetron consists of a solid copper cylindrical cathode (with radius $a$ ) and a surrounding anode (with radius $b$ ). The latter has the shape of a thick cylindrical shell into which cylindrical cavities are drilled. These cavities are known as "resonators". One of the resonators is coupled to an antenna which will transmit the microwave energy out; we will ignore the antenna in the following. All internal spaces are in vacuum. We will consider a typical magnetron with eight resonators, as depicted in Figure 1(a). The three-dimensional structure of a single resonator is shown in Figure 1(b). As indicated there, each of the eight cavities behaves as an inductor-capacitor (LC) resonator, with operating frequency $f=2.45 \mathrm{GHz}$.

A static uniform magnetic field is applied along the magnetron's longitudinal axis, pointing out of the page in Figure 1(a). In addition, a constant voltage is applied between the anode (positive potential) and the cathode (negative potential). Electrons emitted from the cathode reach the anode and charge it, such that they excite an oscillation mode in which the sign of the charge is opposite between every two adjacent resonators. The oscillation of the cavities amplify these oscillations.

The process described above creates an alternating electric field with the aforementioned frequency $f=2.45 \mathrm{GHz}$ (blue lines in Figure 1(a); the static field is not plotted) in the space between the cathode and the anode, in addition to the static field caused by the applied constant voltage. In the steady state, the typical amplitude of the alternating electric field between the anode and the cathode is approximately $\frac{1}{3}$ of the static electric field there. The electron motion in the space between the cathode and the anode is affected by both the static and the alternating parts of the field. This causes electrons that reach the anode to transfer about $80 \%$ of the energy they acquire from the static field into the alternating field. A minority of the ejected electrons returns to the cathode and releases additional electrons, further amplifying the alternating field.

Each resonator can be thought of as a capacitor and an inductor, see Figure 1(b). The capacitance mainly arises from the planar parts of the resonator surface, while the inductance stems from the cylindrical part. Assume that the current in the resonator flows uniformly very close to the surface of its cylindrical cavity, and that the strength of the magnetic field generated by this current is 0.6 times that of an ideal infinite solenoid. The various lengths defining the resonator geometry are given in Figure 1(b). The vacuum permittivity and permeability are $\varepsilon_{0}=8.85 \cdot 10^{-12} \frac{\mathrm{F}}{\mathrm{m}}$ and $\mu_{0}=4 \pi \cdot 10^{-7} \frac{\mathrm{H}}{\mathrm{m}}$, respectively.

Question: A.1 Use the above data to estimate the frequency $f_{\text {est }}$ of a single resonator. (Your result may differ from the actual value, $f=2.45 \mathrm{GHz}$. Use the actual value in the remainder of the question.)

Solution:
The frequency of an LC circuit is $f=\omega / 2 \pi=1 /(2 \pi \sqrt{L C})$. If the total electric current flowing along the boundary of the cavity is $I$, it generates a magnetic field whose magnitude (by the assumptions of the question) is $0.6 \mu_{0} I / h$, and a total magnetic flux equal to $\pi R^{2} \times$ $0.6 \mu_{0} I / h$, hence the inductance of the resonator is $L=0.6 \pi \mu_{0} R^{2} / h$. Approximating the capacitor as a plate capacitor, its capacitance is $C=\varepsilon_{0} l h / d$. Putting everything together, we find

$$
f_{e s t}=\frac{1}{2 \pi} \frac{1}{\sqrt{L C}}=\frac{1}{2 \pi} \sqrt{\frac{h}{0.6 \pi R^{2} \mu_{0}} \frac{d}{\varepsilon_{0} l h}}=\frac{1}{2 \pi} \frac{c}{R} \sqrt{\frac{d}{0.6 \pi l}}=\frac{1}{2 \pi} \frac{3 \cdot 10^{8}}{7 \cdot 10^{-3}} \sqrt{\frac{1}{3.6 \pi}}=2.0 \cdot 10^{9}
$$

$\mathrm{Hz}$

Final answer: $2.0 \times 10^{9}$ Hz