Problem — Field B in the charging capacitor

Problem

A parallel-plate capacitor with circular plates of radius $R=5{,}0$ cm is being charged by a constant current $i=2{,}0$ A; the field $\vv E$ between the plates grows linearly with time. (a) Write the Ampère-Maxwell law along a circle of radius $r<R$ centred on the axis. (b) Derive $\abs{\vv B}=\dfrac{\mu_0 i r}{2\pi R^2}$ and show that at $r=R$ it matches the external field $\dfrac{\mu_0 i}{2\pi R}$. (c) Sketch the trend of $\abs{\vv B}(r)$.

Solution

Setup — displacement current. No charge flows between the plates, but the flux of $\vv E$ changes with time. Maxwell introduces the displacement current $i_s=\varepsilon_0\dfrac{d\Phi_E}{dt}$. Over the whole plate ($A=\pi R^2$) the flux is $\Phi_E=E\,\pi R^2$, so:

$i_s=\varepsilon_0\,\pi R^2\,\frac{dE}{dt}$

Charging the capacitor requires this displacement current to equal the charging current $i$:

$\varepsilon_0\,\pi R^2\,\frac{dE}{dt}=i$

(a) Ampère-Maxwell on the circle $r<R$. On the inner circle we enclose only the fraction of displacement current passing through the area $\pi r^2$, i.e. a fraction $(r/R)^2$ of the total $i$:

$\oint \vv B\cdot d\vv\ell=B\,(2\pi r)=\mu_0\,i\,\frac{r^2}{R^2}$

(b) Internal field and matching. Solving for $B$:

$B\,(2\pi r)=\mu_0\,i\,\frac{r^2}{R^2}$

$\ev{B=\frac{\mu_0\,i\,r}{2\pi R^2}}\qquad (r<R)$

Outside, all the current $i$ is enclosed, as around a wire:

$B=\frac{\mu_0\,i}{2\pi r}\qquad (r>R)$

At $r=R$ the two expressions coincide:

$\frac{\mu_0\,i\,R}{2\pi R^2}=\frac{\mu_0\,i}{2\pi R}=\frac{\mu_0\,i}{2\pi R}$

The field is continuous: it grows linearly from $0$ up to the peak at $r=R$, then decreases as $1/r$.

(c) Graph of $\abs{\vv B}(r)$.

Links

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