Penetration depth of quasistatic H-field into a conductor

Penetration depth of quasi-staticH-field into a conductor

Section 59

Consider a good conductor in anexternal periodic magnetic field

The conductor is penetrated by the H-field,

which induces a variable E-field,

which causes “eddy” currents.

Penetration of field is determined by the thermalconduction equation

Thermometric conductivity

Temperature propagates a distance

in time t.

Quiz: How does the propagation ofheat depend on time?

1.Linearly

2.Quadratic

3.Square root

Since H satisfies the same heat conduction equation

Then H penetrates a conductor to a characteristic depth

Characteristic timethat H has a givenpolarity

(Ignore the factor 2.)

Induced E and eddy currents penetrate to the samedepth

Quiz: How does the skin depthdepend on frequency?

1.Inverse

2.Square root

3.Inverse Square root

A periodic field varies as Exp[-it]

Two limits

1.“low” frequencies

2.“high” frequencies (still below THz)

Low frequency limit

This is the same equation as holds in the static case, when  = 0.

(Periodicfields)

The solution to the static problem is HST(r), which is independent of .

The solution of the slow periodic problem is HST(r)Exp[-it], i.e. the field variesperiodically in time at every point in the conductor with the same frequencyand phase.

H completelypenetrates theconductor

Low frequency recipe

1.Solve for the static H field

2.Multiply by Exp[-it]

3.Find E-field by Faraday’s law

4.Find j by Ohm’s law

In zeroth approximation E = 0 inside conductor

Ohm’s law

Maxwell’s equation

(29.7) for static H-field

E-field and eddy currents appear inside the conductor in the next approximation

The spatialdistribution of E(r)is determined bythe distribution ofthe static solutionHST(r)

Not zero in the nextapproximation

Eddy currents

By Ohm’s law

Equationsfor E inthe lowfrequencylimit

High frequency limit

We are still in the quasi-staticapproximation, which requires

= electron relaxation (collision) time

AND

>> electron mean free path

This means frequencies << THz.

In the high frequency limit

H penetrates only a thin outer layer of the conductor

To find the field outside the conductor, assume exactly

This is the superconductor problem (section 53), where field outside asuperconductor is determined by the conduction B = 0 inside.

Then, to find the field inside the conductor

Consider small regions of the surface to beplanes

The field outside is

What is H0(r) near the surface?

In vacuum, = 1

H0(r) is the solutionto thesuperconductorproblem

To find the field that penetrates, we need to know it just outside, thenuse boundary conditions

In considering B(e)(r), we assumed B(i) = 0.

Since div(B) = 0 always,

The following boundary condition always applies

Just outside the conductor surface in the highfrequency quasi-static case, Bn(e) = 0.

Thus, H0,n = 0, and

H0(r) must be parallel to the surface.

At high frequencies,  ~ 1.

The boundary condition is then

So H on both sides of the surface is

(Parallel to the surface)

A small section of the surface is considered plane,with translational invariance in x,y directions.

Then H = H(z,t)

= 0 (for homogeneous linear medium)

= 0

Since Hz = 0 at z = 0, Hz = 0 everywhere inside.

Hz does not change with z inside.

The equation satisfied by quasi-static H is

Possible solutions of SHO equation are oscillating functions.

This one decreases exponentially with z

This one diverges with z. Discard.

Inside conductor,high  limit

From H inside, we now find E-field inside (high frequency limit)

Phase shift

Magnitudes

Compare vacuum to metal

•For electromagnetic wave in vacuum

•E = H (Gaussian units)

•E and H are in phase

•For high frequency quasi-static limit in metal

Not inphase

Wavelength in metal is  not 

Linearly polarized field:

Phase can be made zero byshifting the origin of time.

Then H0 is real.

Take

Then

But  is also the characteristic damping length.

Not much of a wave!

E and j have the same distribution

E and j lead H by 45 degrees

Quiz: At a given position within a conductor the low frequencylimit, how does the electric field depend on frequency?

1.Increases linearly with f.

2.Increases as the square root of f.

3.Decreases as the inverse square root of f.

At a given position within a conductor the high frequency limit,how does the electric field depend on frequency?

1.Increases as Sqrt[f]

2.Decreases as Exp[-const*Sqrt[f]]

3.Decreases as Sqrt[f] *Exp[-const*Sqrt[f]]

High frequency electric field in a conductor

Complex “surface impedance”of a conductor

Eddy currents dissipate field energy into Joule heat

Heat loss perunit time

Conductorsurface

Mean field energyentering conductor perunit time

Also

We are going to use both of these equationsto find  dependence of Q in the two limits.

Low frequency limit:

~

High frequency limit:

Homework

~

Quiz: In low frequency fields, how does the rate ofJoule heating for a metal depend on frequency?

1.Decreases as inverse square root of f

2.Increases linearly in f

3.Increases as f2

In high frequency fields, how does the rate ofJoule heating for a metal depend on frequency?

1.Increases as square root of f

2.Decreases as 1/f

3.Increases as f2

A conductor acquires a magnetic moment in a periodic externalH-field with the same period.

The change in free energy is due to

1.Dissipation

2.Periodic flow of energy between thebody and the external field

Time averaging the changeleaves just dissipation

Rate of changeof free energy

The mean dissipation of energy per unit time is

Dissipation is determined by the imaginary part of the magnetic polarizability

infrared

Quiz: What is apossible frequencydependence for theelectric field at a givenpoint inside a metal?

.P