(a) Show directly that Eqs. 9.197 satisfy Maxwell’s equations (Eq. 9.177) and the boundary conditions (Eq. 9.175).

(b) Find the charge density, λ(z,t), and the current,I(z,t) , on the inner conductor.

Short Answer

Expert verified

(a) The equation 9.197 satisfies Maxwell’s equation and the boundary conditions.

The charge density is λ(z,t)=2πε0E0cos(kz-ωt),and the current density isI=2πε0μ0Ccos(kz-ωt)

Step by step solution

01

Expression for the electric and magnetic fields at a distance s from the axis of the co-axial transmission line:

Write the expression for the electric and magnetic fields at a distance s from the axis of the co-axial transmission line (Eqs 9.197).

E(s,ϕ,z,t)=Acos(kz-ωt)ss^B(s,ϕ,z,t)=Acos(kz-ωt)csϕ^

Here, E is the electric field, B is the magnetic field, s is the Poynting vector, k is the wave number, c is the speed of light,ϕis the phase andt is the time.

02

Satisfy Maxwell’s equation ∇×E= and ∇×B=0:

(a)

Prove E=0as follow:

·E=1sssEs+1sEϕϕ+Ezz …… (1)

Here, the value of , localid="1657501054364" Es,Eϕand localid="1657501101055" Ezis given as:

localid="1657501166011" Es=Acos(kx-ωt)s

Eϕ=0

Ez=0

Substitute localid="1657500893088" Es=Acos(tα-ωt)s,Eϕ=0andEz=0, in equation (1).

Prove, ·B=0

·B=1sssBs+1sBϕϕ+Bzz …… (2)

Here, the value of localid="1657502944630" Bs,Bϕ,and Bzis given as:

Bs=0Bϕ=Acos(kz-ωt)sBz=0

Substitute Bs=0,Bϕ=Acos(kz-ωt)sandBz=0, in equation (2).

·B=1ss(s×0)+1sAcos(kz-ωt)sϕ+0

B=0

Prove ×E=1sEzϕ-Eϕzs^+Esz-Ezsϕ^+1sssEϕ-Esϕz^

×E=0+Eszϕ^-1sEsϕz^

×E=-E0ksin(kz-ωt)sϕ^×E=-Bt

Prove ×B=1c2Et

×B=1sBzϕ-Bϕzs^+Bsz-Bzsϕ^+1sssBϕ-Bsϕz^×B=-Bϕz+1sssBϕz^×B=E0kcsin(kz-ωt)ss^×B=1c2Et

Satisfy the boundary conditions.

E=Ez=0B=B5=0

Therefore, equation 9.197 satisfies Maxwell’s equation and the boundary conditions.

03

Determine the charge density and current density on the inner conductor:

(b)

Apply Gauss law for a cylinder of radius s and length dzto determine the charge density

E·da=E0cos(kz-ωt)s(2πs)dzE0cos(lz-ωt)s(2πs)dz=Qenclagedε0E0cos(kz-ωt)(2π)dz=λdzε0

Hence, the charge density will be,

λ(z,t)=2πε0E0cos(kz-ωt)

Apply Ampere’s law for a circle of radius s to determine the current density.

B·dl=E0ccos(kz-ωt)s(2πs)1B·dl=μ0ler1dased

Hence, the current density will be,

I=2πε0μ0Ccos(kz-ωt)

Therefore, the charge density is λ(z,t)=2πε0E0cos(kz-ωt), and the current density is I=2πε0μ0ccos(kz-ωt).

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