(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 =

µ0

2π

×

I0r

R2

= 2 × 10−7 ×

6.9√2×10−6×3×10−2

(6×10−2)2

B0 = 1.63 × 10−11 T