Equations of the Quasi-static Field

Post on 22-Feb-2016

42 Views

Category:

Documents

0 Downloads

Preview:

Click to see full reader

DESCRIPTION

Equations of the Quasi-static Field. Section 58. This requires: period of E >> electron response time. Otherwise E changes sign before macroscopic charge flow can build up. How does an E -field appear inside a conductor?. - PowerPoint PPT Presentation

Transcript

Equations of the Quasi-static Field

Section 58

The field at P due to the proximity of the conductor is the same in both cases, except that the magnitude and sign of the field at P in the latter case is changing monotonically as Exp[-i w t]

The equations of the static field should hold at the field point P

Let “l” be the dimension of the conductor

This requires that the oscillation period of E >> electron response time t.Otherwise E changes sign before macroscopic charge flow can build up.

Inside the conductor, the conduction currents are non-zero

These assume that Ohm’s law holds with the DC value of the conductivity: j = s E.

How does an E-field appear inside a conductor?

• Even though dh/dt is small, don’t discard it inside the conductor.

• The external slowly-varying e-field does not penetrate, but the h-field does.

• Thus, dh/dt is the source for a macroscopic E-field inside.

This is the internal E-field that drives the internal currents j

• Requires: electron mean free path << characteristic length for changes in field.

• Otherwise j at given point could be determined by E at distant points.

Local relations

Internal E arises from variations in B. To find B, we need H & constitutive relation.

We get two equations for H, where are enough to find it everywhere at all times

Same as the heat conduction equation (Fourier’s equation):

Thermometric conductivity

Boundary conditions

For a non-ferromagnetic medium, or ferromagnetic one at high frequency, m = 1.Then B = H, and boundary conditions become H1 = H2.

Since j is finite at the interface, see (30.2)

If a conductor has different parts with different s, then H1 = H2 is not enough.Then we also need E1t = E2t at each interface.

Suppose external field is suddenly removed. The induced E and j don’t vanish immediately.Neither does H in or around the conductor.The decay of the field is determined by

Seek solutions of the form

This is an eigenvalue equation. For a given conductor, solutions Hm(x,y,z) that satisfy the boundary conditions exist only for certain gm (real and positive).{Hm(x,y,z)} = complete orthogonal set of eigenvectors.

Let the initial field distribution be

The rate of decay is determined mainly by the smallest gm. Call it g1.

l = dimension of the conductor = largest dimension of the problemGives longest t, smallest g

High conductivity and large conductor give slow decay

Quiz: If the external field is suddenly turned off, the field inside a conductor

1. Turns off at the same time2. Decays exponentially in time with a multiple

time constants3. Decays exponentially in time with a single

time constant4. Decays linearly in time

top related