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# edyn - Physics 54 Time-Varying Fields The only cure for...

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Physics 54 Time-Varying Fields The only cure for boredom is curiosity. There is no cure for curiosity. — Dorothy Parker Completion of Ampere’s Law As an example of a time-varying situation, consider the B-field established by the current charging a capacitor. Shown is a parallel-plate capacitor with circular plates. We are interested in the B-field at point P , shown in the edge-on view. This situation has axial symmetry, which we exploit to use Ampere's law and calculate B . The field lines are circles around the symmetry axis, so we choose as our path of integration a circle through point P . We find B d r = 2 π rB = μ 0 I linked . Here I linked is the current passing through the area bounded by the circular path. But what area? The simplest is a circular disk bounded by the circle of radius r . Because there is no current through such a disk, we would have I linked = 0 , from which we predict B = 0 at P . But experimentally the field is not zero at P . Ampere's law gives a prediction that is wrong. However, we could use a different surface bounded by the circular path. Consider a bowl-shaped surface with the circle as its rim; the bowl surface can be chosen to go around the left plate so that the wire penetrates through it. In that case we would conclude that I linked = I , leading to B = μ 0 I 2 π r . This is in agreement with experiment. These different choices of surface bounded by the path give contradictory results. How do we know which surface to use? Ampere’s law itself gives no instruction on this point. No law of nature can be this vague. It must be either modified or discarded. I a I P r PHY 54 1 Electrodynamics

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Now consider the flux of the E-field through these two surfaces bounded by the circle. First use the circular disk surface. Outside the region between the plates the E-field is negligible, so the flux of E through the disk surface is that through a circle of radius a: Φ e = E d A = π a 2 E = π a 2 Q ε 0 π a 2 = Q ε 0 . (Here we have used E = σ / ε 0 and σ = Q / π a 2 .) From this we find ε 0 d Φ e dt = dQ dt = I . This shows that the quantity ε 0 d Φ e / dt has the dimensions of a current, which in this case is the same as the current in the wires. Maxwell called it the displacement current . The term “displacement” arose from a model Maxwell had in mind, and has no significance today. He suggested that Ampere’s law needed to be completed by adding the displacement current to the “true” current, so that the law reads B d r = μ 0 I linked + ε 0 d dt E d A . Now let us apply this to our capacitor situation to see how it resolves the ambiguity. If we choose the disk surface, then I linked = 0 but ε 0 d dt E d A = I , to we find B = μ 0 I 2 π r , as we should. If we choose the bowl surface we have I linked = I and ε 0 d dt E d A = 0 , again leading to B = μ 0 I 2 π r . The choice of surface no longer matters.
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