equations of the quasi-static field
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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 PresentationTRANSCRIPT
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Equations of the Quasi-static Field
Section 58
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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]
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The equations of the static field should hold at the field point P
Let “l” be the dimension of the conductor
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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.
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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.
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This is the internal E-field that drives the internal currents j
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• 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
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Internal E arises from variations in B. To find B, we need H & constitutive relation.
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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
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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)
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If a conductor has different parts with different s, then H1 = H2 is not enough.Then we also need E1t = E2t at each interface.
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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
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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.
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Let the initial field distribution be
The rate of decay is determined mainly by the smallest gm. Call it g1.
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l = dimension of the conductor = largest dimension of the problemGives longest t, smallest g
High conductivity and large conductor give slow decay
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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