Physics, 03.09.2019 01:10 samarahbrown6050

Use maxwell's equations to show that the direction of the electric field and magnetic field in a plane wave travelling in the x-direction through empty space are perpendicular to the direction of propagation. assume the wave is in a region where no sources are present.

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Answer from: den14

Explanation:

So let's start with Maxwell's equations in wave equation form. Since it's empty space and no sources are present, we can simplify some things:

Assuming simple media we get:

Conduction Current=J=0No charge density $\rho$ =0Permeability ( $\mu$ ) and permittivity ( $\epsilon$ ) are those of empty space ( $\mu_{o}$ , $\epsilon_{o}$ )

We can then start substituting these into Faraday's law or Ampere's Law, let's start with Faraday's just because:

Δ x ( Δ x E = $-\mu_{o}*\frac{\delta H}{\delta t}$ ) removing the parenthesis:

Δ x Δ x E = $-\mu_{o}*\frac{\delta}{\delta t}$ (Δ x H))

We need to sub Ampere's law into this equation.

Ampere's Law: (Δ x H) = $\epsilon_{o}\frac{\delta E}{\delta t}$

With the identity of the cross product of vectors we can solve the left side into:

Δ x Δ x E = Δ (Δ . E)-Δ^{2}*E And we know that (Δ . E)= $\frac{\rho}{\epsilon}$ And since $\rho$ =0 then (Δ . E)=0, making Δ (Δ . E) = 0

The equation then changes to:

Δ^{2}*E= $-\mu_{o}\epsilon_{o}\frac{\delta^{2}E }{\delta t^{2} }$

we also know that the speed of light, c = $\frac{1}{\sqrt{\mu_{o} \epsilon_{o} } }$

The equations end up looking like this:

Δ^{2}*E= $-\frac{1}{c^{2} } \frac{\delta^{2}E }{\delta t^{2} }$

Δ^{2}*H= $\frac{1}{c^{2} } \frac{\delta^{2}H }{\delta t^{2} }$

These are the wave equations (general form)

We then need the Time-Harmonic Maxwell's Equations (this just means that the equations are periodic with respect of time):

$\overline{E}=E(x,y,z)e^{jwt}$

$\overline{H}=H(x,y,z)e^{jwt}$

We then take these two equations and sub them into the wave form equations:

ΔxE=-jw $\mu_{o}$ H

ΔxH=jw $\epsilon_{o}$ E

If you solve the cross product, to the left of both equations, you get:

Δ^{2}*E+ $w^{2} \mu_{o} \epsilon_{o} E$ =0

Δ^{2}*H+ $w^{2} \mu_{o} \epsilon_{o} H$ =0

These are called Helmholtz Equations

Now when we plug in the plane wave solution into these equations, we can finally relate directions of the magnetic and electric field. The Plane Wave Solution equations are:

$E=E_{o}e^{-j(k.r-wt)}$

$H=H_{o}e^{-j(k.r-wt)}$

With k= $(K_{x}, K_{y}, K_{z})$ , this is called the wave vector, and r= (x, y, z) , which is the direction of propagation.

So we take the plane wave solutions and plug them into the Helmholtz Equations, taking into account that the derivatives are spatial derivatives so when we calculate the derivative of wt, it is considered a constant, thus equal to 0, and to simplify, r= $z\overline{z}$ :

For $E=E_{o}e^{-j(k.r-wt)}$ we end up with:

$-k^{2} E_{o}e^{-jkz}+w^{2} \mu_{o} \epsilon_{o}E_{o}e^{-jkz} = 0$

We can then cancel $E_{o} e^{-jkz}$ on both sides:

$k^{2} =w^{2} \mu_{o} \epsilon_{o} =\frac{w^{2} }{c^{2} }$

$k=\frac{w}{c}$ with $k=\frac{2\pi }{\lambda}$ or spatial frequency and $w=\frac{2\pi }{T}$ or temporal frequency

We then move to Gauss' Law with $\rho=0$ :

Δ.E=0

Δ. $E_{o} e^{-jkz} =0$

$E_{o}$ .Δ $e^{-jkz}$ =0

$E_{o} .(-jk)e^{-jkz}=0$

$(E_{o}.k )(-je^{-jkz} )=0$ This is the dot product of $(E_{o} . k)$ and a scalar $-je^{-jkz}$ so:

$(E_{o}.k)=0$ For the dot product to be 0 E and k must be perpendicular

Now for the magnetic field:

ΔxE=-jw $\mu_{o}$ H with $\overline{E}=E_{o}e^{-jkz}$

$\overline{H}=\frac{1}{wave impedance}\overline{k}x\overline{E}$ and $waveimpedance=\frac{w\mu_{o} }{k}$

So the Magnetic field must be orthogonal to the electric field in order for the cross product not be yield a 0 result.

Finally, if E and H are orthogonal to each other and E is perpendicular to the direction of propagation, then H is also perpendicular to the direction of propagation.

Answer from: Quest

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