1. Penetration depth of quasi-static H-field into a conductor Section 59

2. 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.

3. Penetration of field is determined by the thermal conduction equation Thermometric conductivity Temperature propagates a distance in time t.

4. Quiz: How does the propagation of heat depend on time? 1.Linearly 2.Quadratic 3.Square root

5. 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

6. Quiz: How does the skin depth depend on frequency? 1.Inverse 2.Square root 3.Inverse Square root

7. A periodic field varies as Exp[-i  t] Two limits 1.“low” frequencies 2.“high” frequencies (still below THz)

8. Low frequency limit This is the same equation as holds in the static case, when  = 0. (Periodic fields)

9. The solution to the static problem is H ST (r), which is independent of . The solution of the slow periodic problem is H ST (r)Exp[-i  t], 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 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

10. In zeroth approximation E = 0 inside conductor Ohm’s law Maxwell’s equation (29.7) for static H-field

11. 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 H ST (r) Not zero in the next approximation Eddy currents By Ohm’s law Equations for E in the low frequency limit

12. High frequency limit We are still in the quasi-static approximation, which requires = electron relaxation (collision) time AND >> electron mean free path This means frequencies << THz.

13. In the high frequency limit H penetrates only a thin outer layer of the conductor

14. 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.

15. Then, to find the field inside the conductor Consider small regions of the surface to be planes

16. The field outside is What is H 0 (r) near the surface? In vacuum,  = 1 H 0 (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

17. 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, B n (e) = 0. Thus, H 0,n = 0, and H 0 (r) must be parallel to the surface.

18. At high frequencies,  ~ 1. The boundary condition is then So H on both sides of the surface is (Parallel to the surface)

19. 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 H z = 0 at z = 0, H z = 0 everywhere inside. H z does not change with z inside. The equation satisfied by quasi-static H is

20. Possible solutions of SHO equation are oscillating functions. This one decreases exponentially with z This one diverges with z. Discard. Inside conductor, high  limit

21. From H inside, we now find E-field inside (high frequency limit) Phase shift

22. 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

23. Wavelength in metal is  not  Linearly polarized field: Phase can be made zero by shifting the origin of time. Then H 0 is real. Take Then But  is also the characteristic damping length. Not much of a wave!

24. E and j have the same distribution E and j lead H by 45 degrees

25. 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.

26. 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]]

27. High frequency electric field in a conductor Complex “surface impedance” of a conductor

28. 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  dependence of Q in the two limits.

29. Low frequency limit: ~

30. High frequency limit: Homework ~

31. 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 f 2.Increases linearly in f 3.Increases as f 2

32. In high frequency fields, how does the rate of Joule heating for a metal depend on frequency? 1.Increases as square root of f 2.Decreases as 1/f 3.Increases as f 2

33. 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.Dissipation 2.Periodic flow of energy between the body and the external field Time averaging the changeleaves just dissipation Rate of change of free energy

34. The mean dissipation of energy per unit time is

38. Dissipation is determined by the imaginary part of the magnetic polarizability infrared

40. Quiz: What is a possible frequency dependence for the electric field at a given point inside a metal?.P