Electromagnetic Theory (comments to: moloney@nextwork.rose-hulman.edu)     

MAPLE text and input lines

cylinder.ms

This calculates the approximate surface charge on a long thin wire, where the wire radius is small compared to its overall length.

It ultimately determines the charges on each subsection of the wire by solving N equations in N unknowns, by specifying that the potential on the axis of each hollow cylindrically- shaped element be the same V.

Potential due to a charged section of a cylinder when the ends of the cylinder are located at Z1 and Z2

Find the potential at different segment distances away from a charged segment.

Make a function for the potential of the section at "dist" sections away from the charged section with rho="charge".

Set up 4 equations to find the 4 unknown charges on the 4 sections.

Now do the same problem with more segments in the cylinder.

Use a loop to create the equations.

wireskin.ms

PHYSICS RESOURCE PACKETS PROJECT File: wireskin.MS Rose-Hulman Institute of Technology Authors: mjm moloney@nextwork.rose-hulman.edu Software: Maple V Release 3 wireskin.ms

BACKGROUND.

Skin depth effects in a long conducting wire at high frequency are readily modelled using numerical integration. This deals directly with the physics without referring to the formidable properties of bessel functions whose argument is complex. The numerical integration scheme actually generates the kelvin functions (ber, bei) and their derivatives, so their features can be examined if desired.

A wire carrying oscillating current is subject to the well-known 'skin depth' effect of current decreasing from the outside of the wire toward the center at high frequencies. This classic problem involves the complication of a pure imaginary term in a bessel-type differetialequation. Texts treating the problem1,2,3 discuss it in terms of ber and bei ('kelvin') functions, a rather daunting approach for most undergraduates.

Numerical integration solves this problem without any reference to bessel functions. It provides one more illustration of the way computers are gradually altering the way physics problems can be approached by students. A cylindrical wire of radius a carrying a current along its axis (the z-axis) is assumed to have a current density J with only a z component. J is proportional to E, and Ez depends only on the radial coordinate r, in addition to its sinusoidal variation in time [1] Ez(r,t) = E(r) e^(iwt) .

E(r) has real and imaginary parts

[2] E(r) = ER(r) + i EI(r) .

To make clear that both parts of [2] contribute to the overall current density, we write the real part of [1], Re ( Ez(r,t) ) = ER(r) cos(wt) - EI(r) sin(wt) .The magnitude of current density in the wire will then be

| Jz | =s ( ER(r)2 + EI(r)2 )1/2 ,

and its phase will be arctan (EI/ER). It is readily shown1 that E(r) is subject to the following equation in a good conductor ( msw >> mew^2) : r E''(r) + E'(r) = i msw r E(r) .

Since msw has dimensions of inverse length squared, we may define L2 = 1/ msw. A change of variables to x=r/L gives

[3] x E''(x) + E'(x) = i x E(x) .

Using the real and imaginary parts of Er, [3] becomes two coupled equations with real coefficients, ready for numerical integration

[4a] x ER''(x) + ER'(x) = - x EI(x) , and [4b] x EI''(x) + EI'(x) = + x ER(x) .

A Runge-Kutta integration scheme for integrating [4a] and [4b] is discussed below, but may as easily (or more easily) be accomplished in symbolic packages like Maple or Mathematica. (The author's students carried out this calculation using symbolic algebra, and then readily plotted the magnitude of E(r) vs. x. ) As variables in a Runge-Kutta (RK) integration scheme4 we may take

y[1] = ER(x) dy[1]/dx = y[2] y[2] = ER'(x) dy[2]/dx = -y[3] -y[2]/x y[3] = EI(x) dy[2]/dx = y[4] y[4] = EI'(x) dy[2]/dx = +y[1] -y[4]/x

with initial conditions y[1] = 1, y[2] = y[3] = y[4] = 0, dy[1]/dx = dy[2]/dx = dy[3]/dx = 0, and dy[4]/dx = 1. ( Taking EI = 0 on the axis amounts to adjusting the time so that the phase of E(r) is zero on the axis. )The integration proceeds from x=0 on the axis, with a unit-valued pure real E(r), until the surface of the wire is reached at an x value depending on L and the wire radius R. Table I was producedin a few seconds on a 25 Mhz 386 PC, using a 4th order RK scheme with a step size of 5x10-4. It shows how the magnitude grows steadily from the axis to whatever value of x represents the radius of the wire. Note that the phase of E(r) changes, along with its magnitude. ( While attempting to check the results in Table I, it was discovered that ER and EI correspond exactly to the ber and bei 'kelvin' functions listed in ref. 6, p. 430. The 'skin depth' d is conventionally1-3 defined as d= (2)1/2 L = ( 2/ msw)1/2 . This is less natural in the differential equation than L, but is conveniently interpreted as the distance over which field amplitude falls by a factor of 1/e.)In a copper at a frequency of 105 hz, L = 0.147 mm so a wire radius of 0.5 mm equals 3.41 L. (The wire radius then occurs at x=3.41.) At this frequency, Table I shows that the current density at the wire surface is about 2.5 times greater than along the wire axis. At a much lower frequency of 103 hz, however, L=1.47 mm. At this frequency, a 0.5 mm wire radius corresponds to x=0.341, and Table I shows that the current density is nearly constant across the cross-section of the wire.Based on the kelvin function asymptotic forms (ref. 5, p. 381), e +x/(2)1/2 / (2 p x)1/2 ,one can say that when the wire radius is a very large number of 'lengths' L , the current density approaches doubling (increasing by a factor of 2.03) for each additional length L. We see this toward the bottom of Table I, where an increase of 1.05 L nearly doubles the surface electric field.

References

[1] W. T. Scott, The Physics of Elecricity and Magnetism, 2nd Ed. (Wiley, New York, 1966), Sec. 10.8.

[2] M. A. Heald and J. B. Marion, Classical Electromagnetic Radiation, 3rd Ed. (Saunders College Publishing, Fort Worth, 1995) Sec 5.6.

[3] W. R. Smythe, Static and Dynamic Electricity, 3rd Ed. (McGraw-Hill, New York, 1968), Sec 10.04.

[4] W. Press et. al, Numerical Recipes In Pascal (Cambridge U. Press, Cambridge, 1989), Sec. 15.1. [5] M. Abramowitz, and I. Stegun, Handbook of Mathematical Functions (U. S. Government Printing Office, 1966).

The last two columns of Table I refers to total current in the wire. The first of these, labelled 'q' divides the magnitude of the total current by 2 Pi x times the magnitude of the current density at the surface. If the current density at the surface were constant for one length L in from the surface, then zero after that, this ratio would be 1.0 [4]. For wire radii greater than roughly x=1.5,Table I shows it is permissible to think of the current being constant, and confined within a region of thickness L from the surface (the wire acts like a hollow tube of thickness L, as pointed out in ref. 4). For a copper wire of radius 0.5 mm, this condition would be met for frequenciesabove 1.9 MHz. (Halving the wire radius would quadruple this frequency.) The wire resistance above this frequency depends inversely on L, so when the frequency is doubled, the resistance increases by a factor of 21/2.

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Put in pure imaginary term for electric field in highly conducting wire skin depth for conducting wire. integrate out to wire radius, expressed in units of L = (mu sigma omega)^(-1/2);

E(r) = Er + i Ei

x = r/L.

x Er'' + Er' = - Ei and x Ei'' + Ei' =+Er, at x=0, we have Er=1, Ei=0=Er'=Ei'

Solution:

Needed to plot a DE evaluated numerically. with imag part, let y(x) = r(x) + I s(x),
and equate real and imag parts of the de. then

Dop(r(x)) = -b s(x), and
Dop(s(x)) = b r(x).

r(0)=1, r'(0)=0, s(0)=0? s'(0)=0

Substitute in values from problem.