Penetration depth of quasi-static H-field into a conductor

Section 59

Consider a good conductor in an external 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 thermal conduction equation

Thermometric conductivity

Temperature propagates a distance in time t.

Quiz: How does the propagation of heat depend on time?

1. Linearly2. Quadratic3. Square root

Since H satisfies the same heat conduction equation

Then H penetrates a conductor to a characteristic depth

Characteristic time that H has a given polarity

(Ignore the factor 2.)

Induced E and eddy currents penetrate to the same depth

Quiz: How does the skin depth depend on frequency?

1. Inverse2. Square root3. Inverse Square root

A periodic field varies as Exp[-iwt]

Two limits1. “low” frequencies2. “high” frequencies (still below THz)

Low frequency limit

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

(Periodic fields)

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

The solution of the slow periodic problem is HST(r)Exp[-iwt], i.e. the field varies periodically in time at every point in the conductor with the same frequency and phase.

H completely penetrates the conductor

Low frequency recipe1. Solve for the static H field2. Multiply by Exp[-iwt]3. Find E-field by Faraday’s law4. 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 spatial distribution of E(r) is determined by the distribution of the static solution HST(r)

Not zero in the next approximation

Eddy currents

By Ohm’s law

Equations for E in the low frequency limit

High frequency limitWe are still in the quasi-static approximation, 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 a superconductor is determined by the conduction B = 0 inside.

Then, to find the field inside the conductor

Consider small regions of the surface to be planes

The field outside is

What is H0(r) near the surface?

In vacuum, m = 1

H0(r) is the solution to the superconductor problem

To find the field that penetrates, we need to know it just outside, then use 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 high frequency quasi-static case, Bn

(e) = 0.Thus, H0,n = 0, andH0(r) must be parallel to the surface.

At high frequencies, m ~ 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 w 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 in phase

Wavelength in metal is ,d not .l

Linearly polarized field:

Phase can be made zero by shifting the origin of time. Then H0 is real.

Take

Then

But d 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 frequency limit, 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 per unit time

Conductor surface

Mean field energy entering conductor per unit time

Also

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

Low frequency limit:

~

High frequency limit:

Homework

~

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

1. Decreases as inverse square root of f2. Increases linearly in f3. Increases as f2

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

1. Increases as square root of f2. Decreases as 1/f3. Increases as f2

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

The change in free energy is due to 1. Dissipation2. Periodic flow of energy between the

body and the external field

Time averaging the change leaves just dissipation

Rate of change of 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 a possible frequency dependence for the electric field at a given point inside a metal?

.P