Problem — Field B between capacitor plates

Problem

A capacitor has circular plates of radius $R_0 = 5\;\text{cm}$. Between the plates the electric field grows at the rate $\dd E/\dd t = 10^{10}\;\text{V/(m·s)}$. The magnetic field generated by the displacement current, at distance $r$ from the axis (with $r < R_0$), is $B(r) = \mu_0\varepsilon_0\,r\,(\dd E/\dd t)/2$. Calculate it at $r = 1\;\text{cm}$. (Recall $\mu_0\varepsilon_0 = 1/c^2$.)

Solution

Setup. The changing electric field between the plates acts as a displacement current and generates a circular magnetic field. Using $\mu_0\varepsilon_0 = 1/c^2$:

$B(r) = \frac{1}{c^2}\,\frac{r}{2}\,\ddt{E}$

Numerical substitution. With $r = 1\;\text{cm} = 0{,}01\;\text{m}$, $\dd E/\dd t = 10^{10}\;\text{V/(m·s)}$ and $c^2 = 9\cdot 10^{16}\;\text{m}^2/\text{s}^2$:

$B = \frac{1}{9\cdot 10^{16}}\cdot\frac{0{,}01}{2}\cdot 10^{10} \approx 5{,}6\cdot 10^{-10}\;\text{T}$

$\ev{B \approx 5{,}6\cdot 10^{-10}\;\text{T}}$

The magnetic field produced by the displacement current is tiny (fractions of a nanotesla), but its existence is essential: it is the Maxwell term that makes electromagnetic waves possible.

Collegamenti

Argomenti: Onde elettromagnetiche Concetti: Corrente di spostamento · Campo magnetico