A parallel plate capacitor with circular plates of radius $R$ is being charged. The displacement current between the plates is equal to the conduction current in the wires. If the rate of change of electric field is $\frac{dE}{dt}$, what is the magnetic field $B$ at a distance $r<R$ from the axis?

Solution

1. According to the Ampere-Maxwell law, the line integral of the magnetic field around a closed loop is given by $\oint \vec{B}\cdot d\vec{l}={\mu }_0{ϵ}_0\frac{d{\Phi }_E}{dt}$.
2. For a circular path of radius $r<R$, the magnetic field $B$ is uniform, so $\oint \vec{B}\cdot d\vec{l}=B(2\pi r)$.
3. The electric flux through the loop is $$\begin{array}{rl}{\Phi }_E& =EA\\ & =E(\pi r^2)\end{array}$$, so the rate of change of flux is $\frac{d{\Phi }_E}{dt}=\pi r^2\frac{dE}{dt}$.
4. Substituting these into the law gives $B(2\pi r)={\mu }_0{ϵ}_0(\pi r^2)\frac{dE}{dt}$.
5. Solving for $B$ yields $B=\frac{{\mu }_0{ϵ}_{0}r}{2}\frac{dE}{dt}$.
6. Hence the answer is (A).