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 300rads−1.
To find the amplitude of B at a point 3.0 cm from the axis between the plates
Solution:
Radius of plates, R=6cm=6×10−2m
Capacitance of capacitor, C=100μF=100×10−6F=10−4F
Voltage of capacitor, V=230V
Frequency of capacitance, ω=300rad/s
By definition of the displacement current,
Id=ε0dϕEdt⟹Id=ε0ddt(EA)⟹Id=εAdEdt
But E=σε0=Qε0A
Substituting this in the abve equation, we get
Id=ε0Addt(Qε0A)⟹Id=ε0A.1ε0A.dQdt⟹Id=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.Id⟹B=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=Vrms1ωC⟹Irms=Vrms×ω×C⟹Irms=230×300×10−4⟹Irms=69×103×10−4=6.9A
Therefore B=μ0r2πR2√2Irms⟹B=4π×10−7×0.03×√2×6.92π×0.06×0.06⟹B=1.63×10−5T
is the amplitude of B at a point 3.0 cm from the axis between the plates.