1 Current Flow Problems

Transcription

1 Physics 704 Notes Sp 08 Current Flow Problems The current density satisfies the charge conservation equation (notes eqn 7) thusinasteadystate, is solenoidal: + =0 () =0 () In a conducting medium, we may relate to the electric field through the conductivity : = = Φ At a boundary between two different media, integration of equation () over a pillbox straddling the boundary gives ˆ is continuous across the boundary (3) continuity of tangential also gives which in most cases is equivalent to ˆ Φ is continuous across the boundary Φ is continuous across the boundary (4) Notice that the boundary condition on normal (3) implies that norm is NOT continuous, thus there must be charge on the boundary (Notes eqn ). This charge is the source of the fields that direct the current flow.. Example An infinite, plane, conducting sheet with conductivity contains a circular region of a different metal with conductivity radius Current enters the sheet at = flowing in the positive direction ( )= 0ˆ Find the pattern of current flow in the sheet. See Jackson.6 for a discussion of cases where Φis not continuous.

2 Firstnotethatif we expect current to flow inwards through the more conducting circular region, but if we expect current to flow around the more resistive "obstacle" (current follows the path of least resistance). In both of the regions (but not at = ) the potential satisfies Laplace s equation, thus the solution is of the form (Jackson eqn.69, page 76, "orthogonal" notes pg 9). In the inner region we exclude the logarithmic term the negative powers of because they diverge at the origin. Φ ()= ( cos + sin ) = In the outer region, we need a uniform electric field 0 = 0 in the direction to drive the current at infinity. The potential 0 = 0 cos describes this field. This corresponds to the =term in the potential. We exclude the logarithmic term the other positive powers of because they diverge at infinity. Thus Φ ()= 0 cos + ( cos + sin ) = The sum in this expression represents the potential due to the charge on the boundary at = Next we apply the boundary conditions at = Continuity of Φ : = ( cos + sin ) = 0 cos + ( cos + sin ) = Making use of the orthogonality of the cosines sines, we may equate term

3 by term to get = (5) = 0 + (6) = (7) Continuity of = = = Φ : X = ( cos + sin ) = 0 cos + Using orthogonality of the trig functions, we have: X = ( cos + sin ) = (8) = 0 + = (9) = (0) The only solution to equations (5) (8) is = =0 Similarly, from equations (7) (0), = =0 We should have expected this result because the input to the system (the current at infinity) is an =mode. The remaining equations (6) (9) give µ = 0 + = 0 + So thenfrom(6), Thus the potential is 0 0 = + = 0 ( ) + () = 0 + = ( ) + = 0 + Φ ()= 0 + cos = 0 + Φ ()= 0 cos + 0 ( ) cos + 3

4 The current is given by = Φ = 0ˆ + ( ) ³ = 0ˆ ˆ cos + ˆ sin The current for is uniform, is 0 if but 0 if as expected. Outside the circle, ( )forcos positive (positive i.e. to the right of the circle), is positive if Thus current lines converge inward to the circle for negative move back outward for positive Again this is what we expected.. Plotting the flow lines. Remember that we can use a complex potential for -D problems (Lea.4, "orthogonal" notes section ), = Φ+ the imaginary part = constant givesusthefield lines. Here there is an extra subtlety because = Φ so gives the current flow lines. We have, with = cos = Re =Re() ( = Φ + = 0 + if ( ) () + if where = = = (cos sin ) thus, taking the imaginary part of (), we have ( = 0 ³ + sin if sin + ( ) + if If 0 no current flowsthroughthecircleweretrievethesolutioninlea Ch.4.4 for fluid flow around a cylinder. If we retrieve the expected undeviated current flow. For in = sin = 0 + So, since in values of = 0 = max + correspond to current flow lines that pass through the circle. The corresponding valueoutsidethecircleis out = sin = in = + ( ) + + ( ) + + 4

5 Thus out as in + this ratio is if as expected. Plot for = ½ 4 = 3 sin sin + 3 Thus the flow lines are given by: () = 3 4 sin = 3 + p 9 sin 6sin.5.5 circle The line charge density at the circular boundary may be found from the normal component of If is the thickness of the sheet, then ( ) = = 0 = 0 cos + 0 ( ) cos 0 cos cos = ( + ) ( + + ) = µ 0 0 cos + Check the dimensions! is zero at the top bottom of the cylinder where is tangent to the circle, is zero everywhere if = When the uniform field is turned on, it takes a very short time ( ) for the current flow to build up this charge at the boundary, at which point a steady state is achieved. 5