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.
Context 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.)
Context answer:
\boxed{$2.0 \times 10^{9}$}


Extra Supplementary Reading Materials:

Task A. 2 below does not deal with the magnetron itself, but helps to introduce some of the relevant physics. Consider an electron moving in free space under the influence of a uniform electric field directed along the negative $y$ axis, $\vec{E}=-E_{0} \hat{y}$, and a uniform magnetic field directed along the positive $z$ axis, $\vec{B}=B_{0} \hat{z}$ ( $E_{0}$ and $B_{0}$ are positive; $\hat{x}, \hat{y}, \hat{z}$ are unit vectors oriented in the conventional manner). Let us denote the electron velocity at time $t$ by $\vec{u}(t)$. The drift velocity $\vec{u}_{D}$ of the electron is defined as its average velocity. We denote by $m$ and $-e$ the mass and charge of the electron, respectively.

Question: A.2 In each of the following two cases, find $\vec{u}_{D}$ if:

1. at $t=0$ the electron velocity is $\vec{u}(0)=\left(3 E_{0} / B_{0}\right) \hat{x}$
2. at $t=0$ the electron velocity is $\vec{u}(0)=-\left(3 E_{0} / B_{0}\right) \hat{x}$.

Solution:
Denoting the electron velocity by $\vec{u}(t)$, in this case the total force applied on it is

$$
\vec{F}=-e\left(-E_{0} \hat{y}+\vec{u}(t) \times B_{0} \hat{z}\right)
$$

Let us write $\vec{u}(t)=\vec{u}_{D}+\vec{u}^{\prime}(t)$, with $\vec{u}_{D}=\left(-E_{0} / B_{0}\right) \hat{x}$ being the drift velocity of a charged particle in the crossed electric and magnetic fields (the velocity at which the electric and magnetic forces cancel each other exactly). Then $\vec{F}=-e \vec{u}^{\prime}(t) \times B_{0} \hat{z}$. Thus, in a frame moving at the drift velocity $\vec{u}_{D}$, the electron trajectory is a circle with constant-magnitude velocity $u^{\prime}=$ $\left|\vec{u}^{\prime}(t)\right|$, and radius $r=m u^{\prime} / e B_{0}$. In the lab frame this circular motion is superimposed upon the drift at the constant velocity $\vec{u}_{D}$. Hence:

1. For $\vec{u}(0)=\left(3 E_{0} / B_{0}\right) \hat{x}$ we find $u^{\prime}=4 E_{0} / B_{0}$ and $r=4 m E_{0} / e B_{0}^{2}$.
2. For $\vec{u}(0)=-\left(3 E_{0} / B_{0}\right) \hat{x}$ we find $u^{\prime}=2 E_{0} / B_{0}$ and $r=2 m E_{0} / e B_{0}^{2}$.

Final answer: $u^{\prime}=4 E_{0} / B_{0}$,$u^{\prime}=2 E_{0} / B_{0}$