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प्रश्न
A parallel plate capacitor (Figure) 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 ac supply with a (angular) frequency of 300 rad s−1.
- What is the rms value of the conduction current?
- Is the conduction current equal to the displacement current?
- Determine the amplitude of B at a point 3.0 cm from the axis between the plates.
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उत्तर
Radius of each circular plate, R = 6.0 cm = 0.06 m
Capacitance of a parallel plate capacitor, C = 100 pF = 100 × 10−12 F
Supply voltage, V = 230 V
Angular frequency, ω = 300 rad s−1
(a) The rms value of conduction current, I = `"V"/"X"_"C"`
Where,
XC = Capacitive reactance
= `1/(ω"C")`
∴ I = V × ωC
= 230 × 300 × 100 × 10−12
= 6.9 × 10−6 A
= 6.9 μA
Hence, the rms value of the conduction current is 6.9 μA.
(b) Yes, conduction current is equal to displacement current.
(c) Magnetic field is given as:
B = `(μ_0"r")/(2pi"R"^2)"I"_0`
Where,
μ0 = Free space permeability = 4π × 10−7 N A−2
I0 = Maximum value of current = `sqrt2"I"`
r = Distance between the plates from the axis = 3.0 cm = 0.03 m
∴ B = `(4pi xx 10^-7 xx 0.03 xx sqrt2 xx 6.9 xx 10^-6)/(2pi xx (0.06)^2)`
= 1.63 × 10−11 T
Hence, the magnetic field at that point is 1.63 × 10−11 T.
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