Chapter 5 Electromagnetic Waves in Conductors

Figure 5.1 - An Electron in an Electromagnetic Wave

At point P the electric field is pointing up and so the electron is accelerated down, the outcomes of this are

if there are no collisions then the electron moves up and down in response to the electric field (this will be covered in more detail in Section 6)
if collisions are present then the electron loses energy through inelastic collisions and so the wave loses energy. So the flow of energy in the system is
wave® electrons® medium via collisions

5.1 Maxwell's Equations in a Conductor

The assumptions we will be making are

the conductor is a HIL material
that we have a e and a µ
there is a conductivity s present
that r_f=0 (see Problem Sheet 7)

with these assumptions we obtain the following for Maxwell's Equations in a conductor

Ñ .E=0 Ñ .B=0 Ñ×E=-µ

¶H

¶ t

Ñ×H=j_f+e

¶E

¶ t

(5.1)

we can relate j_f to E by using Ohm's Law

j_f=sE (5.2)

If we take the curl of Faraday's Law we get

Ñ× (Ñ×E)=-µ

¶ t

Ñ×H

Ñ (Ñ .E)-Ñ²E=-µ

¶ t

æ
ç
ç
è

¶E

¶ t

+sE

ö
÷
÷
ø

as Ñ .E=0 we get

Ñ²E=µs

¶E

¶ t

+µe

¶²E

¶ t²

(5.3)

This is known asthe Telegraph Equation. A possible solution is (assuming that it is at noramal incidence)

(x,t)=E₀e

(kx-w t)

y (5.4)

by substituting the solution into equation 5.3 we get

-k²E=-wµsE-w²eµE

k²

w²eµ

+1 (5.5)

from this we find that the solution is complex so we set k to be

k=k_r+ k_i

by subsituting this into equation 5.4 we obtain

(x,t)=E₀e

(k_rx-w t)

-k_ix

y (5.6)

where e^(k_^r^{x-w t)} is the oscillatory part of the solution and e^-k_ⁱ^x is the damped part.

5.2 Waves in Good Conductors

By going to the Ampere-Maxwell equation

Ñ×H=sE+e

¶E

¶ t

we can say that a good conductor has

|sE|>>

½
½
½
½

¶E

¶ t

½
½
½
½

s >>we (5.7)

this equation is okay for radiation incident upon a metal for visible light and also for lower frequencies (energies) than visible light. For example for copper at 1Mhz

~ 10¹²

for this equation 5.5 becomes

k²

w²µe

-k=w

sµ

by using de Moivre's theorem

=cos

+sin

therefore

k=w (1+ )

æ
ç
ç
è

sµ

ö
÷
÷
ø

(5.8)

and so

k=(1+ )k_c

where

k_c=

æ
ç
ç
è

wsµ

ö
÷
÷
ø

(k_c>>k_vacuum)

finally

(x,t)=E₀e

-k_cx

(k_cx-w t)

y (5.9)

this is the equation for a wave undergoing damped oscillation. The properties of this equation are

Velocity and Wavelength:

the phase speed is

v_p=

k_r

k_c

µs

(5.10)

where

v_p

2we₀

<<1

this property implies that the wave is dispersive, this means that v_p depends on w .

For example if we take radiation of frequency 1Mhz upon copper and that s_Cu=5.7× 10⁷ Ohm^-1m^-1 and so

v_p=419 ms^-1~ 10^-6c

and

l =0.4× 10^-3 m

(where in a vacuum l~ 300 m )

Attenuation and Skin Depth:

we define skin depth ( d ) as the distance in which the wave amplitude falls to 1/e~ 0.368 of its original amplitude. So equation 5.9 gives

d =

k_c

wµs

(5.11)

For copper at 1Mhz we find

d =66× 10^-6 m

Profile of the Wave:

Figure 5.2 - The Profile of the Wave

Resistivity for High Frequency Waves:

Figure 5.3 - Direct Current in a Wire

by integrating over the cross section we get

j=sE

I=s Ep r²

v=El

I=s v

p r²

R_DC

R_{direct current}=

p r²s

If we now consider the electric field only in the layer d

Figure 5.4 - The Electric Field in the Layer d

again we integrate by using Ohm's Law

ó
õj.dS

ó
õE.dS

I=s E2p rd

R_{high frequency}=

2p rds

if we now use equation 5.11

R_{high frequency}=

2p r

µw

(5.12)

the ratio of this is

R_DC

R_{high frequency}

<<1

Figure 5.5 - Resistivity Due to Frequency

Magnetic Field in a Conductor:

µw

(x×E)

where k=(1+ )k_c and k=x , substituting this into equation 5.9 gives us

(1+)k_c

µw

E₀e

-k_cx

(k_cx-w t)

because 1+ =2e^p/4 we get

µw

E₀e

-k_cx

(k_cx-w t+

)

z (5.13)

and k_c=1/d . This gives a phase difference with respect to the electric field of p/4 .

The impedence

Z_C=

E_y

H_z

µw

Z_c=(1- )

µw

=(1- )

k_c

(5.14)

for example copper at 10GHz gives

|Z|_c=0.036W

which is much less than the impedence in a vacuum

5.3 Reflection at a Dielectric to Metal Interface

Figure 5.6 - Reflection at a Dielectric to Metal Interface

We will assume a normal incidence so that k=kx . The dielectric medium will have a impedence of Z₁ and the metal medium will have an impedence of Z_c .

We use the Fresnel equations (equations 4.15 and 4.17) for when q_I=0

E_R0

E_I0

Z_c-Z₁

Z_c+Z₁

we have

Z_c

Z₁

=(1- )

k_c

Z₁s

=(1- ) a

where a=k_c/Z₁s<<1 as it is a good conductor.

E_R0

E_I0

(1- )a-1

(1- )a+1

if we manipulate this we get

E_R0

E_I0

-1-2 a+2a²

(a+1)²+a²

we can neglect the a² as a<<1 so

E_R0

E_I0

1+2 a

1+2a

~ -1

so upon reflection we find there is a phase shift of p . The reflection co-efficient is defined as

½
½
½
½

E_R0²

E_I0²

½
½
½
½

~ 1-4a

R=1-2

2we

(5.15)

so R~ 1 means a complete reflection back.

For example copper at 10GHz has R~ 1-310^-4 so metal reflects.

We note that there is a frequency dependence of R , so when near the infra-red region R~ 0.95 however in reality R~ 0.99 .