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A parallel plate capacitor made of circular plates each of radius 10.0 cm has a capacitance200muF. The capacitor is connected to a 200V a.c. supply with an angular frequency of 200rads^-1. (a) What is the r.m.s. value of the conduction current? (b) Is the conduction current equal to displacement current? (c) Peak value of displacement current. (d) Determine the amplitude of magnetic field at a point 2.0 cm from the axis between the plates. |
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Answer» Solution :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` |
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