Electromagnetic Fields and Waves: Notes, Solved Problems and Interactive Lab

About the further exercises of Section 7.4

Five further exercises on the circular guide, in the spirit of Problems 7.4.1–7.4.2, the worked examples and the home exercises, graded from easy to harder. They use only the results listed at the head of this subsection. (Five more, on the coaxial line, close Section 7.5.)

Exercise 7.4.4

Wall currents and slots

An air-filled circular guide with $a = 1.0$ cm carries the $cos ϕ$ TE $_{11}$ mode at $f = 11$ GHz.

Find the surface current $J_{S}$ on the wall and the ratio of the largest axial to the largest azimuthal current.
A narrow slot cut along $z$ (axial) or around the pipe (circumferential) does not disturb the mode if no current crosses it. Where can each be cut?
A probe for a slotted measuring section is pushed through an axial slot. Where does it see the strongest field, relative to the field on the axis?
Repeat (b) for TE $_{01}$ and TM $_{01}$ . Why does a TM $_{01}$ rotary joint need a choke?

Try it yourself first.

Strategy

On a perfect conductor $J_{S} = a_{n} \times H$ , with $a_{n} = - a_{ρ}$ the normal pointing from the wall into the guide. Only the tangential $H_{ϕ}$ and $H_{z}$ at $ρ = a$ matter. A slot is harmless when it runs parallel to the current.

At $ρ = a$ , with $J_{1} (x_{11}^{'}) = J_{1} (1.841) = 0.582$ : $H_{z} = C J_{1} (x_{11}^{'}) cos ϕ$ and, from Eq. (7.4.20), $H_{ϕ} = \frac{j τ}{h ^{2} a} C J_{1} (x_{11}^{'}) sin ϕ$ . Then

J_{S} = - a_{ρ} \times (H_{ϕ} a_{ϕ} + H_{z} a_{z}) = H_{z} a_{ϕ} - H_{ϕ} a_{z} = C J_{1} (x_{11}^{'}) [cos ϕ a_{ϕ} - \frac{j τ}{h ^{2} a} sin ϕ a_{z}] .

$f_{c} = 8.79$ GHz, $s = 1 - (8.79/11)^{2} = 0.601$ , $τ = 230.4 \times 0.601 = 138.5$ rad/m, $h = 184.1$ rad/m:

\frac{∣ J _{z} ∣ _{m a x}}{∣ J _{ϕ} ∣ _{m a x}} = \frac{τ}{h ^{2} a} = \frac{138.5}{184. 1 ^{2} \times 0.01} = 0.409.

An axial slot at $ϕ_{0}$ is crossed only by $J_{ϕ} \propto cos ϕ_{0}$ : it is harmless at $ϕ_{0} = 9 0^{\circ}$ and $27 0^{\circ}$ , the top and the bottom, where the central $E$ meets the wall. A circumferential slot at $ϕ_{0}$ is crossed by $J_{z} \propto sin ϕ_{0}$ : a short one is harmless at $ϕ_{0} = 0$ and $18 0^{\circ}$ , the sides. A full ring cuts $J_{z}$ somewhere and always radiates.

The probe couples to the normal field $E_{ρ} = \frac{j ω μ _{0}}{h ^{2} a} C J_{1} (x_{11}^{'}) sin ϕ$ , largest at $ϕ = 9 0^{\circ}$ , exactly where the axial slot is harmless. On the axis $∣ E ∣ = ω μ_{0} ∣ C ∣/ (2 h)$ (Section 7.4.8), so the wall field is

\frac{∣ E _{ρ} ( a , 9 0 ^{\circ} ) ∣}{∣ E ( 0 ) ∣} = \frac{2 J _{1} ( 1.841 )}{1.841} = 0.632

of the central field.

TE $_{01}$ : $H_{ϕ} = 0$ and $H_{z}$ is uniform around the wall, so $J_{S} = H_{z} a_{ϕ}$ runs only around the pipe. Circumferential slots, even full rings, are harmless anywhere, axial slots radiate everywhere. TM $_{01}$ : $H_{z} = 0$ and $H_{ϕ}$ is uniform, so $J_{S} = - H_{ϕ} a_{z}$ is purely axial. Axial slots are harmless, circumferential ones are not. The rotating gap of a rotary joint is a full circumferential slot across the TM $_{01}$ current. A quarter-wave choke presents a short circuit at the gap without metal contact and lets the axial current pass.

Answer

J_{S} = C J_{1} (1.841) [cos ϕ a_{ϕ} - j 0.409 sin ϕ a_{z}] axial slot at 9 0^{\circ} short circumferential slot at 0^{\circ} TE_{01} : rings harmless TM_{01} : axial slots harmless, the gap needs a choke

Check

Charge conservation on the wall, $\nabla \cdot J_{S} = - j ω ρ_{S}$ with $ρ_{S} = - ε_{0} E_{ρ} (a)$ (the normal points along $- a_{ρ}$ ). With $A = C J_{1} (x_{11}^{'})$ and the factor $e^{- j τ z}$ :

\frac{1}{a} \frac{\partial J _{ϕ}}{\partial ϕ} + \frac{\partial J _{z}}{\partial z} = - \frac{A}{a} sin ϕ - \frac{τ ^{2}}{h ^{2} a} A sin ϕ = - \frac{k ^{2}}{h ^{2} a} A sin ϕ, j ω ε_{0} E_{ρ} (a) = j ω ε_{0} \frac{j ω μ _{0}}{h ^{2} a} A sin ϕ = - \frac{k ^{2}}{h ^{2} a} A sin ϕ .

The two sides agree. The charges sit at the top and the bottom, where $E$ ends on the wall.

Exercise 7.4.4. The wall current of the \cos\phi TE_{11} mode at one instant, on the wall cut open along \phi=180^\circ and laid flat, over one guide wavelength (a=1 cm, 11 GHz, \lambda_g=45.4 mm). The current runs around the pipe near \phi=0 and 180^\circ and along it near \pm90^\circ, where it is 0.409 times weaker. The axial slot at 90^\circ and the short circumferential slot at 0^\circ lie along the current and do not disturb the mode. — **Figure 7.4.14.** Exercise 7.4.4. The wall current of the $cos ϕ$ TE $_{11}$ mode at one instant, on the wall cut open along $ϕ = 18 0^{\circ}$ and laid flat, over one guide wavelength ( $a = 1$ cm, $11$ GHz, $λ_{g} = 45.4$ mm). The current runs around the pipe near $ϕ = 0$ and $18 0^{\circ}$ and along it near $\pm 9 0^{\circ}$ , where it is $0.409$ times weaker. The axial slot at $9 0^{\circ}$ and the short circumferential slot at $0^{\circ}$ lie along the current and do not disturb the mode.

Remark

The helix lining of the TE $_{01}$ trunk lines (Application 7.4.2) is this argument made into hardware: closely wound insulated wire is a wall of circumferential conductors with axial gaps everywhere. TE $_{01}$ does not notice, while every mode that needs an axial current sees an open wall and loses its power into the absorbing jacket. The slotted line in part (c) is the circular version of the classic slotted rectangular guide, whose slot runs along the middle of the broad wall for the same reason.

cf. Application 7.4.1

Exercise 7.4.4: Wall currents and slots