GATE 2022 EC Q.52: solution and answer

GATE 2022 · ECQ.52 · 2 marks · MSQ

A waveguide consists of two infinite parallel plates (perfect conductors) at a separation of $1 0^{- 4}$ cm, with air as the dielectric. Assume the speed of light in air to be $3 \times 1 0^{8}$ m/s. The frequency/frequencies of TM waves which can propagate in this waveguide is/are ________.

(A)
$6 \times 1 0^{15}$ Hz
(B)
$0.5 \times 1 0^{12}$ Hz
(C)
$8 \times 1 0^{14}$ Hz
(D)
$1 \times 1 0^{13}$ Hz

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Answer and solution

ANSWER

TM waves propagate only above the TM $_{1}$ cut-off $f_{c, 1} = c / (2 d) = 1.5 \times 1 0^{14} Hz$ , so the correct options are (A) $6 \times 1 0^{15} Hz$ and (C) $8 \times 1 0^{14} Hz$ . The official key is MTA (marks were given to all candidates).

Official answer: marks to all candidatesSource: Official answer key, GATE 2022 EC (IIT Kharagpur)

Plate separation: $d = 1 0^{- 4} cm = 1 0^{- 6} m$ . The dielectric is air, so $c = 3 \times 1 0^{8} m/s$ .
Put the plates at $y = 0$ and $y = d$ . In a TM wave, $E_{z}$ is tangential to both perfectly conducting plates, so it must be zero at $y = 0$ and $y = d$ . That gives $E_{z} \propto sin (\frac{nπ y}{d})$ with $n = 0, 1, 2, \dots$ The phase constant of the TM $_{n}$ mode is $β_{n} = (\frac{2 π f}{c})^{2} - (\frac{nπ}{d})^{2}$ It is real, so the mode propagates, only when $f > f_{c, n} = \frac{n c}{2 d}$ . Below that frequency $β_{n}$ is imaginary and the mode is evanescent.
For $n = 0$ the longitudinal field is $E_{z} = 0$ , so that mode is the TEM mode, not a TM wave. A true TM wave needs $E_{z} \neq = 0$ , so the lowest TM mode is TM $_{1}$ .
Cut-off frequency of TM $_{1}$ : $f_{c, 1} = \frac{c}{2 d} = \frac{3 \times 1 0 ^{8}}{2 \times 1 0 ^{- 6}} = 1.5 \times 1 0^{14} Hz$ Every higher TM mode has a higher cut-off, $f_{c, n} = n f_{c, 1}$ . So a frequency carries some TM wave exactly when $f > 1.5 \times 1 0^{14} Hz$ .
(A) $6 \times 1 0^{15} Hz = 40 \times (1.5 \times 1 0^{14} Hz)$ , which is above $f_{c, 1}$ , so it propagates. TM $_{1}$ through TM $_{39}$ all propagate. (C) $8 \times 1 0^{14} Hz \approx 5.33 \times (1.5 \times 1 0^{14} Hz)$ , which is above $f_{c, 1}$ , so it propagates. TM $_{1}$ through TM $_{5}$ propagate.
(B) $0.5 \times 1 0^{12} Hz$ and (D) $1 \times 1 0^{13} Hz$ are both below $1.5 \times 1 0^{14} Hz$ . No TM mode propagates at these frequencies: all of them are evanescent.
So the frequencies at which TM waves propagate are (A) and (C).
About the official key: the organising institute awarded marks to all candidates for this question and published no reason. One reading changes the answer. Some texts label the TEM mode as "TM $_{0}$ ", with $f_{c} = 0$ . If TM $_{0}$ is counted as a TM wave, every option propagates. Under the usual definition of a TM wave ( $E_{z} \neq = 0$ ), only (A) and (C) propagate.

KEY IDEA

Between parallel plates, the TM $_{n}$ cut-off is $f_{c, n} = n c / (2 d)$ . The $n = 0$ mode has $E_{z} = 0$ , which makes it TEM, so the lowest true TM mode is TM $_{1}$ with $f_{c} = c / (2 d)$ . A TM wave propagates only at frequencies above that cut-off.

WATCH OUT FOR

Leaving the separation in the wrong unit. Taking $d = 1 0^{- 4} m$ instead of $1 0^{- 6} m$ gives $f_{c} = 1.5 \times 1 0^{12} Hz$ , which wrongly adds (D). Two other errors: using $f_{c} = c / d$ instead of $c / (2 d)$ , or counting the TEM ( $n = 0$ ) mode as a TM wave and ticking all four options.

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GATE 2022 EC Q.52: solution and answer

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