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Physics LibreTexts

13.12: Chapter- 12

  • Page ID
    25601
  • Problem (12.1).

    Microwave power of 1 Watt at a frequency of 24 GHz is transmitted through a piece of rectangular waveguide whose inside dimensions are 1 cm x 0.5 cm. Let the z-axis lie parallel with the waveguide axis, and let the microwaves be propagating in the +z direction. Use ε= ε0 and µ= µ0.

    (a) Write expressions for the electric and magnetic fields in the waveguide if the time variation is e-iωt .

    (b) Calculate the amplitudes of the electric and magnetic field components.

    (c) Calculate the time-averaged energy density contained in the fields.

    (d) With what velocity is the above energy density transported along the waveguide?

    (e) Show that the magnetic field vector rotates with time at points which are part way across the width of the waveguide. Show that for points near x=a/4 the rotation is clockwise when viewed from a point on the plus y-axis and looking towards the x-z plane, whereas the rortation is counter-clockwise near x=3a/4.

    Answer (12.1).

    (a) For a frequency F= 24 GHz, ω= 2\(\pi\)F= 1.508 x 1011 radians/sec. For the TE10 mode (all other modes are cut-off)

    \[\mathbf{E}_{\mathbf{y}}=\mathbf{E}_{0} \ \textbf{sin} \left(\frac{\pi \mathbf{x}}{\mathbf{a}}\right) \mathbf{e}^{\mathbf{i}\left(\mathbf{k}_{\mathbf{g}} \mathbf{z}-\omega \mathbf{t}\right)}\nonumber,\]

    where the waveguide walls are at x=0,a and at y=0,b: there is no spatial variation along the narrow dimension of the guide. The field components must satisfy the wave equation: in particular,

    \[\nabla^{2} E_{y}=\varepsilon_{0} \mu_{0} \frac{\partial^{2} E_{y}}{\partial t^{2}}, \nonumber\]

    from which

    \[\left(\frac{\pi}{a}\right)^{2}+k_{g}^{2}=\left(\frac{\omega}{c}\right)^{2}.\nonumber\]

    For the present case, \(\frac{\pi}{a}=314.2 \ \mathrm{m}^{-1}\)

    \[\frac{\omega}{c}=502.7 \ \mathrm{m}^{-1}\nonumber\]

    so that

    \[k_{g}=392.4 \ \mathrm{m}^{-1}.\nonumber\]

    From curlE= iωµ0H, using the fact that E has only a y component, one finds

    \[\mathbf{ {H}_{ {x}}=-\left(\frac{ {k}_{ {g}}}{\omega \mu_{0}}\right) sin \left(\frac{\pi {x}}{ {a}}\right) {E}_{0} {e}^{ {i}\left( {k}_{ {g}} {z}-\omega t\right)},} \nonumber\]

    and \(\mathrm{i} \omega \mu_{0} \mathrm{H}_{\mathrm{z}}=\frac{\partial \mathrm{E}_{\mathrm{y}}}{\partial \mathrm{x}}=\left(\frac{\pi}{\mathrm{a}}\right) \mathrm{E}_{0} \cos \left(\frac{\pi \mathrm{x}}{\mathrm{a}}\right) \mathrm{e}^{\mathrm{i}\left(\mathrm{k}_{\mathrm{g}} \mathrm{z}-\omega \mathrm{t}\right)},\)

    or

    \[\mathbf{H_{z}=\frac{-i \pi}{\mu_{0} a} E_{0} \ \text{cos} \left(\frac{\pi x}{a}\right) e^{i\left(k_{g} z-\omega t\right)}}.\nonumber\]

    Note that \(E_{y}=-Z_{g} H_{x}\) where \(\mathrm{Z}_{\mathrm{g}}=-\left(\frac{\omega}{\mathrm{ck}_{\mathrm{g}}}\right) \mathrm{Z}_{0}\), and Z0= µ0c= 377 Ohms.

    (b) \[\mathrm{S}_{\mathrm{z}}=-\mathrm{E}_{\mathrm{y}} \mathrm{H}_{\mathrm{x}} \text { Watts } / \mathrm{m}^{2}.\nonumber\]

    \[<S_{z}>=-\frac{1}{2} \operatorname{Real}\left(E_{y} H_{x}^{\star}\right)=\frac{1}{2} \frac{\left|E_{0}\right|^{2}}{\left|Z_{g}\right|} \sin ^{2}\left(\frac{\pi{x}}{a}\right).\nonumber\]

    The average across the guide is given by

    \[<<S_{z}>>=\frac{1}{4} \frac{\left|E_{0}\right|^{2}}{\left(\frac{\omega}{c k_{g}}\right) Z_{0}},\nonumber\]

    where E0 is the electric field amplitude. Now \(Z_g=\mathrm{Z}\left(\frac{\omega}{\mathrm{ck}_{\mathrm{g}}}\right)=482.9 \text { Ohms }\), and <<Sz>>ab= 1 Watt, therefore \(<< S_{z}>> = 2 \times 10^{4} \text { Watts } / m^{2}\),

    so that E0 = 6216 Volts/meter, or 31.1 Volts across the narrow dimension of the waveguide. The x-component of the magnetic field amplitude is \(\mathbf{\left|H_{x}\right|=12.87 \ \text { Amps } / m}\). The amplitude of the longitudinal magnetic field component is |H0|= 10.31 Amps/m.

    (c) The time-averaged energy density contained in the fields is given by

    \[<W>=<\varepsilon_{0} E_{Y}^{2} / 2>+<\mu_{0} H_{x}^{2} / 2>+<\mu_{0} H_{z}^{2} / 2>, \nonumber\]

    or

    \[<W>=\frac{\varepsilon_{0} E_{0}^{2} \sin ^{2}(\pi \times / a)}{4}+\frac{1}{4 \mu_{0}}\left(\frac{\mathrm{k}_{\mathrm{g}}^{2}}{\omega^{2}} \mathrm{E}_{0}^{2} \sin ^{2}(\pi \mathrm{x} / \mathrm{a})+\frac{\pi^{2}}{\mathrm{a}^{2} \omega^{2}} \mathrm{E}_{0}^{2} \cos ^{2}(\pi \mathrm{x} / \mathrm{a})\right).\nonumber\]

    Averaged over the guide cross-section, this expression gives

    \[<<\mathrm{W}>>=\varepsilon_{0} \frac{\mathrm{E}_{0}^{2}}{4} \ \text { Joules } / \mathrm{m}^{3}=85.4 \times 10^{-6} \ \mathrm{J} / \mathrm{m}^{3}.\nonumber\]

    (d) The group velocity is the rate of energy transport down the guide;

    \[<<S_{z}>>=V_{g}<< W>>.\nonumber\]

    It follows from this that

    \[V_{g}=c \frac{k_{q}}{(\omega / c)}=0.781 c=\mathbf{2.34 \times 10^{8} \ {m} / \text { sec. }}\nonumber\]

    The group velocity is also given by \(\mathrm{V}_{\mathrm{g}}=\frac{\partial \omega}{\partial \mathrm{k}_{\mathrm{g}}}\).

    (e) Near x=a/4 \(\mathrm{H}_{\mathrm{x}}=\frac{-\mathrm{k}_{\mathrm{g}}}{\mu_{0} \omega} \frac{\mathrm{E}_{0}}{\sqrt{2}} e^{-\mathrm{i} \omega \mathrm{t}}\)

    \[\mathrm{H}_{\mathrm{z}}=\frac{\pi}{\mathrm{a} \mu_{0} \omega} \frac{\mathrm{E}_{0}}{\sqrt{2}} e^{-(\mathrm{i} \omega \mathrm{t}-\pi / 2)}, \nonumber\]

    therefore if \(\mathrm{H}_{\mathrm{x}}=\frac{-\mathrm{k}_{\mathrm{q}}}{\mu_{0} \omega} \frac{\mathrm{E}_{0}}{\sqrt{2}} \cos \omega \mathrm{t}\),

    then

    \[\mathrm{H}_{\mathrm{z}}=-\frac{\pi}{\mathrm{a} \mu_{0} \omega} \frac{\mathrm{E}_{0}}{\sqrt{2}} \sin \omega t.\nonumber\]

    These expressions describe an elliptically polarized wave (nearly circularly polarized because \(\frac{\mathrm{k}_{\mathrm{q}}}{(\pi / \mathrm{a})}=1.25\) rotating in the direction from z to -x, i.e. clockwise looking from +y towards the x-z plane.

    Similarly, near x=3a/4 \(\mathrm{H}_{\mathrm{x}}=-\frac{\mathrm{k}_{\mathrm{g}}}{\mu_{0} \omega} \frac{\mathrm{E}_{0}}{\sqrt{2}} \cos \omega \mathrm{t}\), and

    \(\mathrm{H}_{\mathrm{z}}=\frac{\pi}{\mathrm{a} \mu_{0} \omega} \frac{\mathrm{E}_{0}}{\sqrt{2}} \quad \sin \omega \mathrm{t}\),

    corresponding to a counter-clockwise rotation looking from +y towards the xz plane.

    Problem (12.2).

    An attempt is made to propagate a 10 GHz microwave signal along a rectangular air-filled waveguide whose internal dimensions are 1 cm x 0.50 cm. Use ε0 and µ0 for the dielectric constant and the permeability.

    (a) Write expressions for the electric and magnetic fields associated with the non-propagating TE10 mode.

    (b) Over what distance is the amplitude of the microwave fields attenuated by 1/e?

    (c) Calculate the z-component of the Poynting vector and show that it corresponds to a periodic flow of energy across the waveguide section whose time average is zero.

    Answer (12.2).

    (a) F= 10 GHz ω= 6.28 x 1010 rad./sec. \(\frac{\omega}{c}=2.094 \times 10^{2} \mathrm{m}^{-1}\).

    \(\frac{\pi}{a}=3.141 \times 10^{2} \mathrm{m}^{-1}\).

    For the TE10 mode \(\mathrm{k}_{\mathrm{g}}^{2}+\left(\frac{\pi}{\mathrm{a}}\right)^{2}=\left(\frac{\omega}{\mathrm{c}}\right)^{2}\),

    from which \(k_{g}^{2}=-5.4831 \times 10^{4}\), and kg= ±i 2.342 x 102 m-1,

    a pure imaginary number. Let kg= i\(\alpha\).

    \[E_{y}=E_{0} \sin \left(\frac{\pi x}{a}\right) e^{-\alpha z} e^{-i \omega t}\nonumber\]

    \[\mathrm{H}_{\mathrm{x}}=-\frac{\mathrm{i} \alpha}{\omega \mu_{0}} \mathrm{E}_{0} \sin \left(\frac{\pi \mathrm{x}}{\mathrm{a}}\right) e^{-\alpha \mathrm{z}} e^{-\mathrm{i} \omega t}\nonumber\]

    \[\mathrm{H}_{\mathrm{z}}=-\mathrm{i}\left(\frac{\pi}{\mathrm{a} \omega \mu_{0}}\right) \mathrm{E}_{0} \cos \left(\frac{\pi \mathrm{x}}{\mathrm{a}}\right) e^{-\alpha \mathrm{z}} e^{-\mathrm{i} \omega t}.\nonumber\]

    (b) The attenuation length is \(\frac{1}{\alpha}=\frac{10^{-2}}{2.34}=4.27 \times 10^{-3} \ \text {meters }\), or

    \[1 / \alpha=4.27 \ \mathrm{mm}.\nonumber\]

    (c) Sz = - EyHx, where for this problem

    \[E_{y}=E_{0} \sin \left(\frac{\pi x}{a}\right) e^{-\alpha z} \cos \omega t,\nonumber\]

    and

    \[\mathrm{H}_{\mathrm{x}}=-\left(\frac{\alpha}{\omega \mu_{0}}\right) \mathrm{E}_{0} \sin \left(\frac{\pi \mathrm{x}}{\mathrm{a}}\right) e^{-\alpha \mathrm{z}} \sin \omega \mathrm{t}.\nonumber\]

    Therefore, \(\mathrm{S}_{\mathrm{z}}=-\mathrm{E}_{\mathrm{y}} \mathrm{H}_{\mathrm{x}}=\frac{\alpha}{\omega \mu_{0}} \mathrm{E}_{0}^{2} \sin ^{2}\left(\frac{\pi \mathrm{x}}{\mathrm{a}}\right) \mathrm{e}^{-2 \alpha \mathrm{z}} \sin \omega \mathrm{t} \cos \omega \mathrm{t}\)

    or \({S}_{\mathbf{z}}=1.483 \times 10^{-3} \mathrm{E}_{0}^{2} \sin ^{2}\left(\frac{\pi \mathrm{x}}{\mathrm{a}}\right) \mathrm{e}^{-2 \alpha \mathrm{z}} \sin 2 \omega \mathrm{t}\)

    since \(\sin \omega t \cos \omega t=\frac{1}{2} \sin 2 \omega t\).

    Problem (12.3).

    (a) Design a rectangular air-filled cavity to operate at 24 GHz in the TE103 mode. The cavity is to be constructed from a length of rectangular waveguide whose internal dimensions are 1 x 0.50 cm. Use ε0 and µ0 for the dielectric constant and the permeability.

    (b) Write expressions for the fields in the cavity at resonance.

    Answer (12.3).

    (a) At 24 GHz ω= 1.508 x 1011 rad./sec \(\frac{\omega}{c}=502.7 \ \mathrm{m}^{-1}\).

    For the TE10 mode the guide wave-number can be calculated from

    \[k_{g}^{2}=\left(\frac{\omega}{c}\right)^{2}-\left(\frac{\pi}{a}\right)^{2}\nonumber\]

    where a= 0.01 m is the broad dimension of the guide:

    \[\mathrm{k}_{\mathrm{g}}=3.925 \times 10^{2} \ \mathrm{m}^{-1}.\nonumber\]

    The guide wavelength is \(\lambda_{\mathrm{g}}=2 \pi / \mathrm{k}_{\mathrm{g}}=1.60 \times 10^{-2} \ \mathrm{m}=1.60 \ \mathrm{cm}\). The length of the cavity should be \(\mathrm{L}=\frac{3 \lambda_{\mathrm{g}}}{2}\) for the TE103 mode;

    \[\bf{{L}=2.40 \times 10^{-2} \ {m}=2.40 \ {cm}.}\nonumber\]

    (b) For the forward propagating wave and a TE10 mode

    \[\mathrm{E}_{\mathrm{y}}=\mathrm{E}_{0} \sin \left(\frac{\pi \mathrm{x}}{\mathrm{a}}\right) e^{\mathrm{i} \mathrm{k}_{\mathrm{g}} \mathrm{z}} e^{-\mathrm{i} \omega \mathrm{t}},\nonumber\]

    For the backward propagating wave

    \[E_{y}=E_{0} \sin \left(\frac{\pi x}{a}\right) e^{-i k_{g} z} e^{-i \omega t}.\nonumber\]

    In the cavity one must set up a standing wave along z which has nodes at z=0 and at \(z=L=\frac{3 \lambda_{q}}{2}\); i.e.

    \[E_{y}=E_{0} \sin \left(\frac{\pi x}{a}\right) \sin \left(\frac{n \pi z}{L}\right) \cos \omega t \nonumber.\]

    From this electric field one can calculate the other field components using \(\operatorname{curl} \mathbf{E}=-\mu_{0} \frac{\partial \mathbf{H}}{\partial \mathrm{t}}\). For the TE10 mode the electric field has only one component, Ey, and

    \[\frac{\partial \mathrm{E}_{\mathrm{y}}}{\partial \mathrm{z}}=\mu_{0} \frac{\partial \mathrm{H}_{\mathrm{x}}}{\partial \mathrm{t}} \quad \quad \quad \quad (1) \nonumber\]

    \[\frac{\partial \mathrm{E}_{\mathrm{y}}}{\partial \mathrm{x}}=-\mu_{0} \frac{\partial \mathrm{H}_{\mathrm{z}}}{\partial \mathrm{t}} \quad \quad \quad \quad (2)\nonumber\]

    From (1)

    \[\mathrm{H}_{\mathrm{x}}=\left(\frac{1}{\mu_{0} \omega}\right)\left(\frac{\mathrm{n} \pi}{\mathrm{L}}\right) \mathrm{E}_{0} \sin \left(\frac{\pi \mathrm{x}}{\mathrm{a}}\right) \cos \left(\frac{\mathrm{n} \pi \mathrm{z}}{\mathrm{L}}\right) \sin \omega t\nonumber\]

    From (2)

    \[\mathrm{H}_{\mathrm{z}}=-\left(\frac{1}{\mu_{0} \omega}\right)\left(\frac{\pi}{\mathrm{a}}\right) \mathrm{E}_{0} \cos \left(\frac{\pi \mathrm{x}}{\mathrm{a}}\right) \sin \left(\frac{\mathrm{n} \pi \mathrm{z}}{\mathrm{L}}\right) \text { sinwt }.\nonumber\]

    For resonance kg= 3λ/2 and therefore L= 2.40 cm.

    Problem (12.4).

    A rectangular waveguide is filled with material characterized by a relative dielectric constant εr= 9.00. The inside dimensions of the waveguide are a= 1 cm, b= 0.50 cm.

    (a) Over what frequency interval would this guide support only the TE10 mode?

    (b) Calculate the time-averaged energy density for the TE10 mode, and average the resulting expression over the guide cross section. Let the amplitude of the electric field be Ey = E0.

    (c) Calculate the time-averaged value of the Poynting vector, and average the resulting expression over the guide cross section. Let the amplitude of the electric field be Ey = E0.

    (d) A signal having an average power of 1 Watt is transmitted down the guide at a frequency of 7.5 GHz. Calculate (i) the wavelength along the guide, λg; (ii) the ratio of the guide wavelength to the free space wavelength for a 7.5 GHz plane wave; (iii) the group velocity, i.e. the velocity with which information can be transmitted down the guide; (iv) the amplitude of the electric field.

    Answer (12.4).

    (a) For the TE10 mode the fields have the form

    \[\mathrm{E}_{\mathrm{y}}=\mathrm{E}_{0} \sin \left(\frac{\pi \mathrm{x}}{\mathrm{a}}\right) \mathrm{e}^{\mathrm{i}\left(\mathrm{k}_{\mathrm{g}} \mathrm{z}-\omega \mathrm{t}\right)}, \nonumber\]

    \[\mathrm{H}_{\mathrm{x}}=-\left(\frac{\mathrm{k}_{\mathrm{g}}}{\omega \mu_{0}}\right) \mathrm{E}_{0} \sin \left(\frac{\pi \mathrm{x}}{\mathrm{a}}\right) \mathrm{e}^{\mathrm{i}\left(\mathrm{k}_{\mathrm{g}} \mathrm{z}-\omega \mathrm{t}\right)}, \nonumber\]

    \[\mathrm{H}_{\mathrm{z}}=-\left(\frac{\mathrm{i}}{\omega \mu_{0}}\right)\left(\frac{\pi}{\mathrm{a}}\right) \mathrm{E}_{0} \cos \left(\frac{\pi \mathrm{x}}{\mathrm{a}}\right) \mathrm{e}^{\mathrm{i}\left(\mathrm{k}_{\mathrm{g}} \mathrm{z}-\omega \mathrm{t}\right)}, \nonumber\]

    where \(\omega^{2} \varepsilon \mu_{0}=k_{g}^{2}+\left(\frac{\pi}{a}\right)^{2}\)

    or \(\varepsilon_{r}\left(\frac{\omega}{c}\right)^{2}=k_{g}^{2}+\left(\frac{\pi}{a}\right)^{2}\).

    If a= 1 cm = 0.01 m \(\left(\frac{\pi}{a}\right)^{2}=9.870 \times 10^{4} \ \mathrm{m}^{-2}\).

    The cut-off frequency corresponds to kg = 0; i.e. \(\sqrt{\varepsilon_{\mathrm{r}}}\left(\frac{\omega}{\mathrm{c}}\right)=\frac{\pi}{\mathrm{a}}\). At cut-off \(\frac{\omega}{\mathrm{c}}=\frac{314 \cdot 2}{\sqrt{\varepsilon_{\mathrm{r}}}}=104.7 \ \mathrm{m}^{-1}\),

    or F = 5.00 GHz.

    For the higher order modes, cut-off corresponds to the condition kg= 0, so that

    \[\varepsilon_{r}\left(\frac{\omega}{c}\right)^{2}=\left(\frac{m \pi}{a}\right)^{2}+\left(\frac{n \pi}{b}\right)^{2}\nonumber,\]

    where \(\frac{\pi}{a}=314.2 \ \mathrm{m}^{-1}\), and \(\frac{\pi}{b}=628.4 \ \mathrm{m}^{-1}\).

    For m=0 n=1 F01 = 10.00 GHz

    m=1 n=1 F11 = 11.18 Ghz

    m=1 n=2 F12 = 20.62 GHz

    m=2 n=0 F20 = 10.00 GHz.

    This waveguide will support only the TE10 mode for frequencies in the interval 5.00 to 10.00 GHz.

    (b) The time-averaged energy density is given by

    \[<W>=<\varepsilon E_{y}^{2} / 2>+<\mu_{0} H_{x}^{2} / 2>+<\mu_{0} H_{z}^{2} / 2>, \nonumber\]

    \[< W>=\frac{\varepsilon_{r} \varepsilon_{0}}{4} E_{0}^{2} \sin ^{2}\left(\frac{\pi_{x}}{a}\right)++\frac{1}{4 \mu_{0} \omega^{2}} k_{g}^{2} E_{0}^{2} \sin ^{2}\left(\frac{\pi_{\mathrm{x}}}{a}\right)+\frac{1}{4 \mu_{0} \omega^{2}}\left(\frac{\pi}{a}\right)^{2} E_{0}^{2} \cos ^{2}\left(\frac{\pi_{\mathrm{x}}}{a}\right).\nonumber\]

    Take the spatial average over the cross-section of the waveguide:

    \[<< W>>=\left(\varepsilon_{r}+\frac{1}{\omega^{2} \varepsilon_{0} \mu_{0}}\left(k_{g}^{2}+\left(\frac{\pi}{a}\right)^{2}\right)\right) \frac{\varepsilon_{0} E_{0}^{2}}{8},\nonumber\]

    \[\mathbf{<< W>>=\frac{\varepsilon_{\mathrm{r}} \varepsilon_{0}}{4} \mathrm{E}_{0}^{2} \ \text{ Joules } / {m}^{3}}.\nonumber.\]

    (c) Sz = - EyHx,

    \[\left\langle\mathbf{S}_{\mathbf{z}}\right\rangle=\frac{\mathbf{k}_{\mathbf{g}}}{\mathbf{2} \omega \mu_{0}} \mathbf{E}_{0}^{2} \sin ^{2}\left(\frac{\pi \mathbf{x}}{\mathbf{a}}\right).\nonumber\]

    The average over the x co-ordinate gives

    \[\mathbf{<< {S}_{ {z}}>>=\frac{ {k}_{ {g}}}{4 \omega \mu_{0}} {E}_{0}^{2} \quad \text { Watts } / {m}^{2}}.\nonumber\]

    (d) The group velocity is such that <<Sz>>= Vg<<W>>, therefore

    \[V_{g}=\frac{c}{\varepsilon_{r}}\left(\frac{k_{q}}{(\omega / c)}\right).\nonumber\]

    At 7.5 GHz \(\mathrm{k}_{0}=\frac{\omega}{\mathrm{c}}=157.1 \ \mathrm{m}^{-1}\) and the free space wavelength is λ0= 4.00 cm. The waveguide wave-vector is given by

    \[k_{g}^{2}=9 k_{0}^{2}-\left(\frac{\pi}{a}\right)^{2}=12.337 \times 10^{4}, \nonumber\]

    and

    \[\mathrm{k}_{\mathrm{g}}=3.513 \times 10^{2} \ \mathrm{m}^{-1}.\nonumber\]

    From this, the guide wavelength is

    (i) \(\mathbf{\lambda_{g}=\frac{2 \pi}{k_{g}}=1.788 \ {cm}},\), and

    (ii) \(\frac{\lambda_{g}}{\lambda_{0}}=0.447\)

    (iii) \(V_{g}=\frac{c}{9}\left(\frac{3.512}{1.571}\right)= \bf{0.745 \times 10^{8} \text { meters/sec }}.\)

    (iv) \(\left\langle\left\langle S_{z}\right\rangle\right\rangle=\frac{k_{q}}{4 \omega \mu_{0}} E_{0}^{2}=\frac{1}{a b}=2 \times 10^{4} \ \text { Watts } / m^{2}\).

    From this \(\mathrm{E}_{0}^{2}=4 \frac{(\omega / \mathrm{c})}{\mathrm{k}_{\mathrm{g}}} (377)\left(2 \times 10^{4}\right)=1.349 \times 10^{7}\),

    so that E0 = 3673 Volts/m.

    Problem (12.5).

    It is desired to construct a cylindrical air-filled cavity which will resonate at 10 GHz in the TE01 doughnut mode (this is a very low loss mode which is often used to construct frequency meters). If the radius of the cavity is chosen to be R= 2.50 cm how long should the cavity be made?

    Answer (12.5).

    For the TE01 mode the tangential component of the electric field, Eθ, is proportional to the Bessel function \(J_{0}^{\prime}\left(\mathrm{k}_{\mathrm{c}} \mathrm{r}\right)=-\mathrm{J}_{1}\left(\mathrm{k}_{\mathrm{c}} \mathrm{r}\right)\) where

    \[\mathrm{k}_{\mathrm{c}}^{2}=\varepsilon_{\mathrm{r}}\left(\frac{\omega}{\mathrm{c}}\right)^{2}-\mathrm{k}_{\mathrm{g}}^{2}, \nonumber\]

    see eqn.(10.90b).

    The component Eθ must be zero at the waveguide wall in order that the tangential component of the electric field be zero:

    \[J_{1}\left(k_{c} R\right)=0\nonumber\]

    or \(\mathrm{k}_{\mathrm{c}} \mathrm{R}=3.8317\) for the lowest mode.

    Thus \(k_{c}=\frac{3.832}{0.025}=153.3 \ \mathrm{m}^{-1}\).

    For an air-filled waveguide εr= 1, so

    \(k_{g}^{2}=2.0373 \times 10^{4} \ \mathrm{m}^{-2}\) since \(\frac{\omega}{\mathrm{c}}=209.44 \ \mathrm{m}^{-1}\) at 10 GHz. Consequently, kg = 142.7 m-1 and the guide wavelength is \(\lambda_{\mathrm{g}}=\frac{2 \pi}{\mathrm{k}_{\mathrm{g}}}=4.40 \ \mathrm{cm}\). But Eθ must vanish at the cavity end walls and therefore Eθ must be proportional to \(\sin \left(\frac{\mathrm{n} \pi \mathrm{z}}{\mathrm{L}}\right)\). Thus \(\mathrm{k}_{\mathrm{g}}=\frac{\mathrm{n} \pi}{\mathrm{L}}\) and the cavity length must be an integral number of half wavelengths long. A convenient choice would be L= 4.40 cm.