The magnetic field inside a charging capacitor

If Maxwell’s addition really has physical meaning, a real magnetic field must exist between the capacitor’s plates while it charges. Let us compute its value exploiting the cylindrical symmetry.

Take two circular plates of radius $R_0$ separated by a small distance: between them the electric field is uniform, directed from one plate to the other, with magnitude $E(t)$. Choose as the path a circle of radius $r$ centred on the axis and parallel to the plates. By rotational symmetry, the magnitude of $\vec{B}$ on that circle is constant and tangent, so $\Gamma_B = B\cdot 2\pi r$. The electric flux linked by the circle is $\Phi_E = E\cdot\pi r^2$ as long as $r < R_0$.

Applying Ampère-Maxwell — with no current of charges inside the capacitor, $i_\text{link} = 0$:

$B(r)\cdot 2\pi r = \mu_0\varepsilon_0\,\frac{d\Phi_E}{dt} = \mu_0\varepsilon_0\,\pi r^2\,\frac{dE}{dt}$

Simplifying:

Field inside ( $r \le R_0$)

$\ev{B(r) = \frac{\mu_0\varepsilon_0\,r}{2}\,\frac{dE}{dt}}$

The magnetic field grows linearly with the distance from the axis, exactly as inside a wire carrying uniform current. Outside the plates ($r > R_0$) the linked electric flux stays capped at $E\cdot\pi R_0^2$, so

$B(r) = \frac{\mu_0\varepsilon_0\,R_0^2}{2r}\,\frac{dE}{dt} \qquad (r > R_0)$

exactly as the magnetic field outside a wire carrying the equivalent current $i = \varepsilon_0\,\pi R_0^2\,dE/dt$.

Perfect symmetry

For someone observing the circuit from outside, a real wire cannot be distinguished from the “fictitious wire” of displacement current: both produce the same magnetic field $B \propto 1/r$.

Order of magnitude

For an equivalent current $i\sim 1$ A, at $r = 1$ cm from the axis: $B \sim \mu_0 i\,r/(2\pi R_0^2)$, of the order of a few $\mu$T — weak, but measurable.

Links

Topics: Onde elettromagnetiche Concepts: Corrente di spostamento · Teorema di Ampère · Campo magnetico Skills: Impostazione simbolica

Related exercises: Problem — Field B in a charging capacitor · Problem — Continuity of B · Problem — Why the displacement current