A parallel plate capacitor (Fig.) 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.

Given, a parallel plate capacitor made of circular plates.

(a) Here,
Radius of the circular plate, R = 6.0 cm
 $$\begin{gathered} \mathrm{Capacitance} \mathrm{of} \mathrm{the} \mathrm{plates}, \mathrm{C} = 100 \mathrm{\rho F} = 100 \times {10}^{-12}\mathrm{F} \\ \mathrm{Angular} \mathrm{frequency}, \mathrm{\omega } = 300 \mathrm{rad} {\mathrm{s}}^{-1} \end{gathered}$$

${\mathrm{E}}_{\mathrm{rms}} = 230 \mathrm{V}$

Therefore, 

           $\boxed{{\mathrm{I}}_{\mathrm{rms}} = \frac{{\mathrm{E}}_{\mathrm{rms}}}{{\mathrm{X}}_{\mathrm{C}}}} = \frac{{\mathrm{E}}_{\mathrm{rms}}}{{\displaystyle \frac{1}{\mathrm{\omega C}}}} = {\mathrm{E}}_{\mathrm{rms}} \times \mathrm{\omega C}$ 

$\therefore$        ${\mathrm{I}}_{\mathrm{rms}} = 230 \times 300 \times 100 \times {10}^{-12}$
                 $= 6.9 \times {10}^{-6}\mathrm{A} = 6.9 \mathrm{\mu A}$ 

(b) Yes, the conduction current is equal to the displacement current.

$$\begin{gathered} \mathrm{I} = {\mathrm{I}}_{\mathrm{D}}, \mathrm{whether} \mathrm{conduction} \mathrm{current}, \mathrm{I} \mathrm{is} \mathrm{steady} \mathrm{d}.\mathrm{c}. \mathrm{or} \mathrm{a}.\mathrm{c}. \\ \mathrm{This} \mathrm{can} \mathrm{be} \mathrm{shown} \mathrm{below} : \end{gathered}$$

          $\boxed{{\mathrm{I}}_{\mathrm{D}} = {\mathrm{\epsilon }}_0\frac{\mathrm{d}\left({\mathrm{\varphi }}_{\mathrm{E}}\right)}{\mathrm{dt}}} = {\mathrm{\epsilon }}_0 \frac{\mathrm{d}}{\mathrm{dt}}\left(\mathrm{EA}\right)$     $\left(\therefore {\mathrm{\varphi }}_{\mathrm{E}} = \mathrm{EA}\right)$

$\Rightarrow$      ${\mathrm{I}}_{\mathrm{D}} = {\mathrm{\epsilon }}_0\mathrm{A}\frac{\mathrm{dE}}{\mathrm{dt}}$
            $= {\mathrm{\epsilon }}_0\mathrm{A}\frac{\mathrm{d}}{\mathrm{dt}}\left(\frac{\mathrm{Q}}{{\mathrm{\epsilon }}_0\mathrm{A}}\right)$      $\left(\because \mathrm{E} = \frac{\mathrm{\sigma }}{{\mathrm{\epsilon }}_0} = \frac{\mathrm{Q}}{{\mathrm{\epsilon }}_0\mathrm{A}}\right)$ 
$\Rightarrow$     $$\begin{gathered} {\mathrm{I}}_{\mathrm{D}} = {\mathrm{\epsilon }}_0\mathrm{A} \times \frac{1}{{\mathrm{\epsilon }}_0\mathrm{A}}\frac{\mathrm{dQ}}{\mathrm{dt}} \\ = \frac{\mathrm{dQ}}{\mathrm{dt}} = \mathrm{I} \end{gathered}$$
(c) We know that, 

                      $\boxed{\mathrm{B} = \frac{{\mathrm{\mu }}_0}{2\mathrm{\pi }}\frac{\mathrm{r}}{{\mathrm{R}}^2}{\mathrm{I}}_{\mathrm{D}}}$ 

This formula goes through even if displacement current, I_D (and therefore magnetic field B) oscillates in time. The formula above shows that they oscillate in phase.
Since I_D = I, we have
                      $\mathrm{B} = \frac{{\mathrm{\mu }}_0\mathrm{rI}}{2{\mathrm{\pi R}}^2}$ 

If I = I₀, the maximum value of current, then
Amplitude of B = maximum value of B 

$$\begin{gathered} = \frac{{\mathrm{\mu }}_0{\mathrm{rI}}_0}{2{\mathrm{\pi R}}^2} = \frac{{\mathrm{\mu }}_0\mathrm{r}\sqrt{2}{\mathrm{I}}_{\mathrm{rms}}}{2{\mathrm{\pi R}}^2} \left(\because {\mathrm{I}}_0 = \sqrt{2} {\mathrm{I}}_{\mathrm{rms}}\right) \\ = \frac{4\mathrm{\pi } \times {10}^{-7} \times 0.03 \times \sqrt{2} \times 6.9 \times {10}^{-6}}{2 \times 3.14 \times (0.06{)}^2}\mathrm{T} \\ = 1.63 \times {10}^{-11}\mathrm{T}. \end{gathered}$$