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Confirm that the energy in theTEmnmode travels at the group velocity. [Hint: Find the time-averaged Poynting vector <S>and the energy density <u>(use Prob. 9.12 if you wish). Integrate over the cross-section of the waveguide to get the energy per unit time and per unit length carried by the wave, and take their ratio.]

Short Answer

Expert verified

Answer

The velocity of energy carried by the electromagnetic wave is confirmed to be the group velocity of the wave.

Step by step solution

01

Expression for the velocity of the energy carried by the electromagnetic wave traveling in a waveguide:

Write the expression for the velocity of the energy carried by the electromagnetic wave traveling in a waveguide.

V=<S>dA<u>dA …… (1)

Here, v is the velocity of electromagnetic wave, Sis the time-averaged pointing vector for the electromagnetic wave, uis the energy per unit volume of the electromagnetic wave, and Ais the cross-section of the waveguide.

Write the expression for the time-averaged pointing vector for the electromagnetic wave.

S=12μ0ReE×B …… (2)

Write the expression for the energy per unit volume of the electromagnetic wave.

u=14Reε0EE+1μ0BB …… (3)

Here, *represents the complex conjugate of the quantity.

02

Determine the time-averaged pointing vector for the electromagnetic wave:

Write the value of E×B.

(B×B)=(Bz*Ey)x^-(Bz*Ex)x^+(Bz*Ex-Bz*Ey)z^

As (Bz*Ey)and (Bz*Ex)are imaginary components, the value of localid="1658406246663" Re(E×B)will be,

localid="1658406255214" Re(E×B)=(Bz*Ex-Bz*Ey)z^

Using equation 9.180 and 9.186, write the value of By*,Ex,Bx*and Ey.

By*=-ik(ωc)2-k2(-nπb)B0cos(mπxa)sin(nπyb)Ex=-(ωc)2-k2(-b)B0cos(mπxa)sin(nπyb)Bx*=-ik(ωc)2-k2(a)B0sin(mπxa)cos(nπyb)Ey=-ik(ωc)2-k2(-a)B0sin(mπxa)cos(nπyb)

Substitute all the above values in equation (2).

<S>=12μ0(By*Ex-Bx*Ey)z^<S>=ωkπ2B02(ωc)2-k2[nb2cos2mπxasin2nπyb+ma2sin2mπxacos2nπybz^]<S>·da=18μ0ωkπ2B02(ωc)2-k2ab[ma2+nb2]<S>·da=18μ0ωkc2B02ab(ωmn2)

03

Determine the energy per unit volume of the electromagnetic wave:

Using the equation9.176, write the value of Eand B.

E=E0ei(kz-ωt)B*=B*0ei(kz-ωt)

Similarly, using equation 9.180 and 9.186, write the value of Bz*and Ez.

Bz*=B0cos(mπxa)cos(nπyb)Ez=0

Substitute the value of localid="1658406358582" Eand localid="1658406367803" Bin equation (3).

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04

Determine the group velocity:


Substitute <S>·da=18μ0ωkc2B02(ωmn2)and <u>·da=B02abω28μ0ωmn2in equation (1).

V18μ0ωkc2B02ab(ωmn3)B02abω28μ0ωmn2V=18μ0ωkc2B02ab(ωmn2)×8μ0ωmn2B02abωV=kc2ωV=cωω2-ωmn2

Therefore, the velocity of energy carried by the electromagnetic wave is confirmed to be the group velocity of the wave.

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