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A parallel plate capacitor (Fig 8.02] made of circular plates each of radius R = 6.0 cm has a capacitance C = 100 pF. The capacitor is connected to a 230 V a.c. supply with a (angular) frequency of "300 rad s"^(-1). (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 vecB at a point 3.0 cm from the axis between the plates. |
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Answer» Solution :(a) Given that radius of each circular plate `R=6.0cm = 0,06m, C =100pH=10^(-10)F, V_("rms")=230V` and angular frequency `omega="300 rad s"^(-1)`. `therefore"rms value of conduction current "I_("rms")=(V_("rms"))/(X_(c))=V_("rms")xxComega` `=230xx10^(-10)xx300=6.9xx10^(-6)A=6.9muA` (B) Yes, the condution current is equal to the displacement current even for the a.c. circuit. (c) As per modified form of Ampere.s circuital law `B=(mu_(0))/(2pi)(I+I_(d)).(r)/(R^(2))`. For a point situated at a distance `r=3.0cm=0.03m` from the AXIS between the plates `I=0`. Hence, `B=(mu_(0))/(2pi)i_(d).(r)/(R^(2))` and the AMPLITUDE of `vecB=B_(0)=(mu_(0)r)/(2piR^(2)).I_("max")=(mu_(0)r)/(2piR^(2)).sqrt2I_("rms")` `therefore""B_(0)=(4pixx10^(-7)xx0.03xxsqrt2xx6.9xx10^(-6))/(2pixx(0.06)^(2))=1.63xx10^(-11)T.` |
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