Displacement Current and Ampere's Law

Q) A circular parallel plate capacitor is charging. What is the magnetic both everywhere between the plates, either within or outside the radius of the plates?

The radius of the plates is R and the rate of change of the field is a constant given as (DE/Dt). Show that:

i) when r0/2)ε0(DE/Dt)r

ii) when r>R B(r) = (μ0/2)ε0(DE/Dt)R2/r

i) To find the magnetic field at r within the capacitor, we are interested in the displacement current that flows through a circle of radius r - this is given by the change in flux through this circle:
ID0(dΦE/dt)]

ID0πr2(DE/Dt)

Symmetry requires that the magnetic field is constant and tangential around the radius r circle. So, by Ampere's law (the only current through our loop is the Displacement current):
B(r)*2πr = μ0ID

B(r) = (μ0/2πr)ε0(DE/Dt)\$#960;r2

B(r) = (μ0/2)ε0(DE/Dt)r

- QED

ii) Outside the capacitor we proceed identically except, the displacement current through the circle is constant, because the electric field is (to a good approximation) only changing in the space between the plates:

ID0πR2(DE/Dt)

B(r) = (μ0/2πr)ε0(DE/Dt)\$#960;R2

B(r) = (μ0/2)ε0(DE/Dt)R2/r

- QED