A parallel capacitor made of circular plates radius 10.0 cm has a capacitance of 200 pF. The capacitor is connected to a 200 a.c. supply with an angular frequency of `200 "rad"" "s^(-1).`
(i) What is the r.m.s. value of conduction current
(ii) Is the conduction current equal to displacement current?
(iii) Find peak value of displacement current ?
(iv) Determine the amplitude of magnetic field at a point 2.0cm from the axis between the plates.

Explanation

Here, `R=10cm=0.1m`,
`C=200pF=200xx10^-12F=2xx10^-10F,`
`E_(rms)=200V, omega=200rads^-1,`
`r=2.0xx10^-2m`.
(a) `I_(rms)=(E_(rms))/(1//omegaC)=omegaCE_(rms)`
`=8xx10^-6A=8muA`
(b) Yes, because `I_D=I`
(c) `I_0=sqrt2I_(rms)=sqrt2xx8xx10^-6`
`=11.312xx10^-6 A`
(d) Consider a loop of radius r between two
circular plates of parallel plate capacitor placed
coaxially with them. The area of this loop `A'=pir^2`.
By symmetry, the magnetic field `vecB` is equal in
magnitude and is tangentially to the circle at every
point, In this case, only a part of displacement current
`I_D` will cross the loop of area A'. Therefore, the
current passing through the area A'. Therefore, the current passing through the area A'
`I'=(I_D)/(piR^2)xxpir^2=(I_D)/(R^2)r^2`
Using Ampere's Maxwell law we have,
`oint vecB.vec(dl)=mu_0xx`(total current through the area A')
or `2pirB=mu_0(I_0)/(R^2)r^2`
or `B=(mu_0I_0r)/(2piR^2)=(4pixx10^-7xx11.312xx10^-6 xx2xx10^-2)/(2pi(0.1)^2)`
`=4.525xx10^-12T`