A parallel plate capacitor made of circular plates each of radius `R= 6.0` cm has a capacitance `c = 100pF`. The capacitor is connected to a `230 V AC` supply with a ( angular) frequency of `300 rad//s`
(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.
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Explanation

(a) Capacitive reactance,
`X_C =1/(omegaC)`
`= 1/( 300 xx 100 xx 10^-12)`
`= (10^8)/3 Omega`
There is only capacitance in the circuit.
`:. i_(r m s) = V_(r m s)/(X_C)`
`=(230)/((10^8//3))`
`=6.9 xx 10^-6 A`
(b) Yes the derivation in example `29.1` is true even if current is alternating.
(c) Here, `i_(d)` is displacement current and `i` the conduction current. Magnetic field at a distance `r` from the axis
`B = (mu_0)/(2 pi) (i_d)/(R^2) r`
`B_(rms) =(mu_0)/(2 pi) i_(rms)/(R^2) r" "(i_(d)=i=i_(rms))`
Substituting the values, we have
`B_(rms) = ((2 xx 10^-7)(6.9 xx 10^-6))/((6 xx 10^-2)^2)(3 xx 10^-2)`
`= 1.15 xx 10^-11 T`
`:. B_0 = sqrt (2) B_(rms)`
`=(sqrt 2)(1.15 xx 10^-11) T`
`= 1.63 xx 10^-11 T`.