## Wednesday, March 23, 2011

### Introducing Basic Electrodynamics (2)

We left off in the last instalment with the two plane wave equations:

E= E(max) cos (kx - wt)

B = B(max) cos (kx - wt)

Taking partial derivatives in each case, the first with respect to x, the second with respect to t:

@E/@x= -k E(max) sin (kx - wt)

-@B/@t= -w B(max) sin (kx - wt)

Since: @E/@x = -@B/@t then:

-kE(max) = -w B(max) or, E(max)/B(max)= w/k

Also, w/k = (2 πf/ 2 πf/L) = c = E/B (where L is the wavelength of the EM wave)

that is, the ratio of the field maximum amplitudes is equal to the ratio of the angular frequency (w = 2 πf) to the wave number vector, k(2 πf/L ). Bear in mind the units of E are in V/m (volts per meter) and for B, Tesla.

Energy Carried by E-M Waves:

The energy carried by an electromagnetic (E-M) wave with field intensities E, B is given by the Poynting vector, S:

S = 1/u_o [E X B]

where again, u_o denotes the magnetic permeability, or u_o = 4 π x 10^-7 H/m.

Physically, the Poynting vector is the rate at which energy flows through a unit surface perpendicular to the flow.

Recall now, for any two vectors A, B

A X B = A B sin Θ

where Θ is the angle between them. If then A is perpendicular to B, then:

sin Θ = sin (90) = 1 so that:

A B sin Θ = AB

By the same token, taking the vectors E, B perpendicular to each other, we find:

S = EB/ u_o

which has units of watts/m^2

Now, recall from earlier, E/B = c, so:

B = E/c

Then: S = EB/ u_o = (E/c) E/u_o = E^2/ c u_o

or: S = (c/u_o) B

The "time average" is also of interest and entails taking the time average of the function: cos^2(kx - wt) (why?)

Which yields:

T(av) {cos^2(kx - wt) } = ½

The average value of S (or Intensity) can then be obtained from the maximum vector amplitudes, viz.

I = S(av) = E(max) B(max)/ 2 u_o

or:

S(av) = E(max)^2/ 2 u_o(c) = cB(max)^2/ 2 u_o

Note that u_o c is a very important quantity known as the "impedance of free space", or:

u_o c = [u_o/e_o]^½

u_o c = [4 π x 10^-7 H/m/ 8.85 x 10^-12 F/m]^½ = 377 ohms

The respective contribution of the two field energies (associated with the electric intensity, E, and magnetic intensity B) can easily be shown to be:

U(E) = ½ e_oE^2

U(M)= ½ (B^2/ u_o)

In a given volume the energy is equally shared by the two fields such that:

U(E) = U(M) = ½ e_oE^2 = (B^2/2 u_o)

The total, instantaneous energy density of the fields is then:

U = U(E) + U(M) = 2(½ e_oE^2 ) = e_oE^2 = B^2/u_o

Averaged over one or more cylces this leads to the total average energy per unit volume:

U(av) = [e_oE^2]av = ½ e_oE(max)^2= B(max)^2/2 u_o

The intensity of an EM wave is then:

I = S (av) = c U(av) = PA

where P denotes the radiation pressure, or P = S/c (for complete absorption) or P = 2S/c for complete reflection of the wave.

(In direct sunlight, one finds P_R = 5 x 10^-6 N/m^2)

Some Problems:

1) An E-M wave in a vacuum has an electric field amplitude E = 220 V/m. Compute the magnitude of the corresponding magnetic field, B.

Solution:

We have: B = E/c = 220 V/m/ (3 x 10^8 m/s) = 7.3 x 10^-7 T

2) An isotropic electromagnetic wave source has a power of 100 watts. At what distance from the wave will the maximum electric intensity E(max) be 15 V/m?

Recall: S(av) = E(max)^2/ 2 u_o(c)

But we define S(av) = Power/area, where in this case Power = 100w

or:

Power = S(av) area

for an isotropic volume or space for which all distances are equal we use a sphere of radius r, and we know the area of a sphere is:

A= 4 π r^2

Then:

Power = S(av) 4 π r^2 = {E(max)^2/ 2 u_o(c)} 4 π r^2

We obviously need to solve for r, so use algebra to re-arrange and obtain:

r = [100w (u_o) c/ 2π E(max)^2 ]^½

substituting all values:

r = [100w (4 π x 10^-7 H/m) (3 x 10^8 m/s) / 2π (15 V/m)^2 ]^½

r = 5.1 m

3) If the amplitude of the magnetic field in an EM wave is B(max)= 4.1 x 10^-8T, then what is the average intensity of the wave?

We have: I = S (av) = c U(av)

but:

U(av) = B(max)^2/2 u_o

therefore:

I = S (av) = c B(max)^2/2 u_o

I - (3 x 10^8 m/s)(4.1 x 10^-8 T)^2 / 2(4 π x 10^-7 H/m) = 0.201 W/m^2