(a) Capacity of capacitor C = 100 pF Capacitive reactance XC =
1
ωC
=
1
300×100×10−12
∴ XC =
108
3
Ω If Irms is the rms value of conduction current
Irms =
Erms
XC
= 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.
∫
→
B
.
→
dl
= µ0[
1
πR2
πr2] B × 2πr =
µ0Ir2
R2
For amplitude of magnetic field, we require I0 = Irms√2 = 6.9 √2 µA So, B0 =