Problem — Field B in a charging capacitor

Problem

A parallel-plate capacitor has circular plates of radius $R=5$ cm and is being charged by a current $I=0{,}5$ A. Calculate the magnitude $\abs{\vv B}$ of the magnetic field at a distance $r=3$ cm from the axis, using $B=\dfrac{\mu_0\,r\,I}{2\pi R^2}$.

Solution

Origin of the field. No real current flows between the plates, but the electric field varies as the capacitor charges: the displacement current generates a circular magnetic field, given (for $r<R$) by $B=\frac{\mu_0\,r\,I}{2\pi R^2}$

Unit conversion. In metres: $r=3\ \text{cm}=0{,}03$ m and $R=5\ \text{cm}=0{,}05$ m.

Substitution with $\mu_0=4\pi\cdot 10^{-7}$: $B=\frac{4\pi\cdot 10^{-7}\cdot 0{,}03\cdot 0{,}5}{2\pi\cdot (0{,}05)^2}$

Simplification ($4\pi/2\pi=2$ and $(0{,}05)^2=2{,}5\cdot 10^{-3}$): $B=\frac{2\cdot 10^{-7}\cdot 0{,}03\cdot 0{,}5}{2{,}5\cdot 10^{-3}}=\frac{3{,}0\cdot 10^{-9}}{2{,}5\cdot 10^{-3}}\approx 1{,}2\cdot 10^{-6}\ \text{T}$

$\ev{B\approx 1{,}2\cdot 10^{-6}\ \text{T}=1{,}2\ \mu\text{T}}$

Connections

Topics: Electromagnetic waves Concepts: Displacement current · Magnetic field