(a) Capacity of capacitor C = 100 pF Capacitive reactance $X_C$ = $\frac{1}{\omega C}$ = $\frac{1}{300 \times 100 \times 10^{-12}}$ ∴ $X_C$ = $\frac{10^8}{3}$ Ω If $I_{\text{rms}}$ is the rms value of conduction current $I_{\text{rms}}$ = $\frac{E_{\text{rms}}}{X_C}$ = 230 × 3 × $10^{-8}$ = 690 × $10^{-8}$ = 6.9 µA (b) Yes, the conduction current in wires is always equal to displacement current within plates. (c) To find magnetic field B of a point 3 cm from the axis within plates, let us assume a loop of radius 3 cm with center on axis. Now modified Ampere’s Law. ∫ $\vec{B} \cdot \vec{dl}$ = $\mu_0 \left[ \frac{1}{\pi R^2} \pi r^2 \right]$ B × 2πr = $\frac{\mu_0 I r^2}{R^2}$ For amplitude of magnetic field, we require $I_0$ = $I_{\text{rms}} \sqrt{2}$ = 6.9 $\sqrt{2}$ µA So, $B_0$ = $\frac{\mu_0}{2\pi} \times \frac{I_0 r}{R^2}$ = 2 × $10^{-7}$ × $\frac{6.9 \sqrt{2} \times 10^{-6} \times 3 \times 10^{-2}}{(6 \times 10^{-2})^2}$ $B_0$ = 1.63 × $10^{-11}$ T