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Question

A parallel plate capacitor (fig.) made of circular plates each of radius R=6.0 cm has a capacitance C=100μF. The capacitor is connected to a 230 V ac supply with a (angular) frequency of 300 rad s1. Determine the amplitude of B at a point 3.0 cm from the axis between the plates.?

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Solution

Given: A parallel plate capacitor made of circular plates each of radius R=6.0 cm has a capacitance C=100μF. The capacitor is connected to a 230 V ac supply with a (angular) frequency of 300rads1.
To find the amplitude of B at a point 3.0 cm from the axis between the plates
Solution:
Radius of plates, R=6cm=6×102m
Capacitance of capacitor, C=100μF=100×106F=104F
Voltage of capacitor, V=230V
Frequency of capacitance, ω=300rad/s
By definition of the displacement current,
Id=ε0dϕEdtId=ε0ddt(EA)Id=εAdEdt
But E=σε0=Qε0A
Substituting this in the abve equation, we get
Id=ε0Addt(Qε0A)Id=ε0A.1ε0A.dQdtId=dQdt=I
The distance of point from the axis between the plates
r=3cm=0.03m
Radius of the plates, R=6cm=0.06m
The magnetic field at a point between the plates
B=μ02πR2.r.IdB=μ0rI2πR2
If I=I0, maximum value of the current then
I=2Irms
and
The rms value of the current
Irms=VrmsXC...........(i)
And XC=1ωC
Substituting this in eqn(i), we get
Irms=Vrms1ωCIrms=Vrms×ω×CIrms=230×300×104Irms=69×103×104=6.9A
Therefore B=μ0r2πR22IrmsB=4π×107×0.03×2×6.92π×0.06×0.06B=1.63×105T

is the amplitude of B at a point 3.0 cm from the axis between the plates.

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