Show that the magnetic field $B$ at a point in between the plates of a parallel-plate capacitor during charging is $B = \frac{{\mu _0 \epsilon _0 r}}{2} \cdot \frac{{dE}}{{dt}}$ (symbols have their usual meanings).

(N/A) Consider a circular loop of radius $r$ between the plates of a parallel-plate capacitor. According to the Ampere-Maxwell law,the line integral of the magnetic field around this loop is given by:
$\oint B \cdot dl = \mu_0 I_d$
$B(2\pi r) = \mu_0 \epsilon_0 \frac{d\phi_E}{dt}$
Since the electric field $E$ is uniform between the plates,the electric flux $\phi_E$ through the circular loop of radius $r$ is $\phi_E = E \cdot A = E(\pi r^2)$.
Substituting this into the equation:
$B(2\pi r) = \mu_0 \epsilon_0 \frac{d}{dt}(E \pi r^2)$
$B(2\pi r) = \mu_0 \epsilon_0 \pi r^2 \frac{dE}{dt}$
Dividing both sides by $2\pi r$,we get:
$B = \frac{\mu_0 \epsilon_0 r}{2} \cdot \frac{dE}{dt}$