Answer to Question #153812 in Field Theory for Sneha

Question

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A rectangular metal waveguide filled with a dielectric of relative permittivity εr = 4, has the inside dimensions 3 x 1.2 cm, the cut off frequency for the dominant mode is

Answer 1

Answer

$V_p = \frac{c }{\sqrt{(μ_rε_r)} }\\here \\μ_r =1 ; ε_r = 4 (given). \\So \\f_c =\frac{V_p}{2a} \\f_c=\frac{3\times10^8}{((3\times10^{-2})\sqrt{(1\times4}) }\\f_c=2.5GHz$