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Temka [501]
3 years ago
11

List, in ascending order, the cutoff frequencies for the first ten modes of a rectangular waveguide, normalized to the cutoff fr

equency of the TE10 mode. (That is, for each mode, you will be calculating the cutoff frequency of that mode divided by the cutoff frequency of the TE10 mode.) Assume that a = 2b. Give numerical values for the normalized cutoff frequencies and clearly indentify the mode number and type (TM or TE) for each mode. (Note that some modes may have the same cutoff frequency.)
Engineering
1 answer:
Alisiya [41]3 years ago
5 0

Answer:

Fcte10 < Fcte01 = Fcte0 < Fctm11 < Fcte21 = Fctm21 < Fcte12 = Fctm12 < Fcte22

Explanation:

Assuming a = 2b

Attached below is the required steps to the solution

The cutoff frequencies for the first ten modes of a rectangular waveguide listed in ascending order  is :

Fcte10 < Fcte01 = Fcte0 < Fctm11 < Fcte21 = Fctm21 < Fcte12 = Fctm12 < Fcte22

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Explanation:

a)

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V(r)=\frac{kq}{r}

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r is the distance from the charge

In this problem, we have a system of two charges, so the total potential at a certain point will be given by the algebraic sum of the two potentials.

Charge 1 is

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q_2=-3.00 \mu C=-3.00\cdot 10^{-6}C

and is located at (x=0, y = 0.40 m)

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r_{2A}=\sqrt{0.40^2+0.40^2}=0.566 m

So the potential due to charge 1 is

V_2=\frac{(8.99\cdot 10^9)(-3.00\cdot 10^{-6})}{0.566}=-4.77\cdot 10^4 V

Therefore, the net potential at point A is

V_A=V_1+V_2=+4.50\cdot 10^4 - 4.77\cdot 10^4=-2700 V

b)

Here we have to calculate the net potential at point B, located at

(x = 0.40 m, y = 0.30 m)

The distance of charge 1 from point B is

r_{1B}=\sqrt{(0.40)^2+(0.30)^2}=0.50 m

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W=(-1.6\cdot 10^{-19})(-26,800)=4.3\cdot 10^{-15} J

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