Question

Question 8.2: A parallel plate capacitor (Fig. 8.7) 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.

Class 12 - Physics - Electromagnetic Waves Electromagnetic Waves Page-286

Answer 1

Given,
radius of each circular plate , R = 6cm = 0.06 m
capacitance of parallel plate capacitor , C = 100pF = 100 × 10^-12 F = 10^-10 F
angular frequency , $\omega$=300rad/s

(a) RMS value of conduction current, $I=\frac{V}{X_C}$
where, $X_C$=capacitative reactance.
$X_C=\frac{1}{\omega C}$

$I=V\omega C=230V\times300rad/s\times10^{-10}F\\I=6.9\times10^{-6}A$

(b) yes, conduction current is equal to displacement current.

(c) magnetic field is given as, $B=\frac{\mu_0r}{2πR^2}I_0$
here, $I_0$ is the maximum current = √2I
r is the distance between the plates from the axis.e.g., r = 3cm = 0.03

so, B = (4π × 10^-7 × 0.03)/(2π × 0.06²) × √2 × 6.9 × 10^-6 T
= 1.63 × 10^-11 T
hence, magnetic field is 1.63 × 10^-11 T