A parallel plate capacitor made of circular plates each of radius 10 cm has a capacitance `200pF`. The capacitor is connected to a 230V a.c. supply with an angular frequency of `400rads^-1`.
(i) What is the r.m.s. value of the conduction current? (ii) Find the amplitude of `vecB` at a point 2.0 cm from the axis of the plates.

Explanation

Here, `R=10cm=0.10m`,
`C=200pF=200xx10^-12F`,
`omega=400rads^-1, V_(rms)=230V`.
(i) `I_(rms)=(V_(rms))/(X_C)=(V_(rms))/(1//omegaC)=V_(rms)xxomegaC`
`=230xx400xx(200xx10^-12)`
`=18.4xx10^-6A=18.4muA`
(ii) Magnetic field at a distance r from the axis
plates is
`B=(mu_0)/(2pi) r/(R^2)I`
Amplitude of `vecB` is given by
`B_0=(mu_0)/(2pi) r/(R^2)I_0=(mu_0)/(2pi) r/(R^2)I_(rms) sqrt2`
`[:' I_(rms)=I_0//sqrt2]`
Here, `r=2.0cm=2.0xx10^-2m, R=0.10m`
`:. B_0=((4pixx10^-7))/(2pi)xx(2.0xx10^-2)/((0.10)^2xx(18.4xx10^-6)`
`=1.04xx10^-11T`