## Friday, March 6, 2015

### Introducing Electrodynamics (2)

1. 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/mo  [E X B]

where again, mo denotes the magnetic permeability, or mo = 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. We already saw it, at least those who worked Problem (2) did, from the previous installment. This was given as: P = E y  H z / c  =   1300 Wm-2 .

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 / m

which has units of Wm-2

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

B = E /c

Then: S = EB m = (E/c) E/ m =   E 2 / c m

B =  mo H

or: S = (c /mo) 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) {cos2 (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 mo

or:

S(av) = E(max) 2/ 2 mo c  = cB(max)2 / 2 mo

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

mo c  =    Ömo  / Ö εo

mo  c  = 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 εE2

U(M)= ½ (B2/ mo)

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

U(E) = U(M) = ½ εo E2 = (B2 /2 mo)

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

U = U(E) + U(M) = 2(½ εo E2) = εE2 = B2 /mo

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

U(av) = [εE2 ]av = ½ εo E(max)2 = B(max)2 /2 mo

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/m2)

2. THE HALL EFFECT

Before getting into Hall current electrodynamics, we begin with some unit vector basics. Along each axes one can define unit vectors: x^, y^ and z^. Then the respective multiplications (by vector directions) yield: x^*y^ = z^, and y^*z^ = x^ and x^*z^ = y^. These rules will always apply for a right handed coordinate system.

Thus, a vector cross product given by: A X B, must always have the directions attached by means of vectors, e.g.:

A(x^) x B(y^) = (A X B) (z^) = C(z^)

Now, as depicted in the accompanying diagram below, consider a slab of conducting material through which a current I flows as shown (in direction x^) so that I = I(x^).

We consider in turn the effect on positive and negative charge carriers, after having attached the coordinate system x, y, z and thence we specify (in addition to the current I(x^):

The magnetic induction: B = B(y^)

The velocity: v = v(-x^)

(Since the velocity of actual charges is opposite to the direction of conventional current flow)

The magnetic force (F = qvB) acting on a unit (+) charge deflects it toward the upper face, resulting in the accumulation of + charges there, and negative (-) charges on the bottom face.

Expressing the force with appropriate directions:

F(z^) = q v(-x^) x B(y^)

The opposite accumulation of charge (+ to bottom, - to top) gives rise to an electrical force that counteracts the magnetic. Eventually equilibrium occurs when:

Eq = qvB

At this point:

E = VH  t

Where VH  is the Hall potential difference.

Then:

(VH / t) q = qvB

Or, by directions:

qE(-z^) = q vB(z^)

or VH = Bvt

The drift velocity can be found from the basic definition of the current:

I = ne v A

Where A is the area A = Lw (length x width of box)

n = number density of charges (per cubic meter)

e = unit of electronic charge = 1.6 x 10-19  C

Solving for v:

V = I / neLw

Therefore, the Hall potential difference is:

VH = B{I/neLw} t = BI/ new

Example Problem:

If the magnetic induction B = 1.0 T, and a rectangular slab of material (such as shown) is for copper, with n = 10^29 /m3, find the Hall emf if I = 10A, and the width of the slab is 0.001 m.

VH = BI/ new

= (1.0T) (10 A)/ {1029 /m3)(1.6 x 10-19 C) (0.001m)}

VH = 0.6 mV

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.

2) A radio wave transmits 25 W/m 2 of power per unit area. A plane surface of area 2.4 m x 0.7 m is perpendicular to the direction of propagation of the wave. Calculate the radiation pressure    P R  on the surface if it is assumed to be a perfect absorber.

3) An AM radio station broadcast isotropically with an average power of 4 kW. A dipole receiving antenna 65 cm long is located 4 miles from the transmitter. Find the Emf induced by this signal between the ends of the receiving antenna.

4) A community plans to build a solar power conversion station, i.e. to convert solar radiation into electrical power. They require 1 MW (megawatt) of power, and the final system is assumed to have an efficiency of 30% (30% of the solar energy incident on the surface is converted to electrical energy). What must be the effective area, A, of an assumed perfectly absorbing surface to be used in such an installation? Assume a constant solar energy flux incident of 1000 W/m2.

5)  The diagram for this problem (left) shows a slab of silver with dimensions: z1 = 2 cm, y1 = 1mm, carrying 200 A of current in the +x^ direction. The uniform B-field has a magnitude of 1.5 T. If there are 7.4 x 1028 free electrons per cubic meter. Find:

a)The electron drift velocity

b)The magnitude and direction of the E-field due to the Hall Effect

c)The magnitude of the Hall EMF.