$A$ parallel plate capacitor (Figure) made of circular plates each of radius $R=6.0\; cm$ has a capacitance $C=100\; pF$. The capacitor is connected to a $230\; V$ ac supply with a (angular) frequency of $300 \;rad \;s ^{-1}$.
$(a)$ What is the rms value of the conduction current?
$(b)$ Is the conduction current equal to the displacement current?
$(c)$ Determine the amplitude of $B$ at a point $3.0\; cm$ from the axis between the plates.

(N/A) Radius of each circular plate,$R = 6.0 \; cm = 0.06 \; m$
Capacitance of a parallel plate capacitor,$C = 100 \; pF = 100 \times 10^{-12} \; F$
Supply voltage,$V = 230 \; V$
Angular frequency,$\omega = 300 \; rad \; s^{-1}$
$(a)$ The rms value of conduction current is given by $I = \frac{V}{X_C}$,where $X_C = \frac{1}{\omega C}$ is the capacitive reactance.
Therefore,$I = V \omega C = 230 \times 300 \times 100 \times 10^{-12} \; A = 6.9 \times 10^{-6} \; A = 6.9 \; \mu A$.
$(b)$ Yes,the conduction current is equal to the displacement current in the circuit.
$(c)$ The magnetic field $B$ at a distance $r$ from the axis between the plates is given by $B = \frac{\mu_0 r}{2 \pi R^2} I_0$,where $I_0 = \sqrt{2} I$ is the peak current.
Given $r = 3.0 \; cm = 0.03 \; m$ and $\mu_0 = 4 \pi \times 10^{-7} \; T \; m \; A^{-1}$.
$B = \frac{4 \pi \times 10^{-7} \times 0.03 \times \sqrt{2} \times 6.9 \times 10^{-6}}{2 \pi \times (0.06)^2} \approx 1.63 \times 10^{-11} \; T$.