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A parallel-plate capacitor has circular plates each of radius R = 6.0 cm and capacitance C = 100 pF. It is connected to a 230 V ac supply of angular frequency 300 rad/s. (a) Find the rms value of the conduction current. (b) Is the conduction current equal to the displacement current? (c) Find the amplitude of the magnetic field B at a point 3.0 cm from the axis, between the plates.

1. I_rms = 6.9 microA; yes (equal); B0 = 1.63*10⁻¹¹ T
2. I_rms = 6.9 mA; no; B0 = 1.63*10⁻⁸ T
3. I_rms = 0.69 microA; yes; B0 = 0.82*10⁻¹¹ T
4. I_rms = 69 microA; no; B0 = 3.26*10⁻¹¹ T

Correct answer: I_rms = 6.9 microA; yes (equal); B0 = 1.63*10⁻¹¹ T

Solution

This is the standard NCERT problem (C = 100 pF). (a) Xc = 1/(omega C) = 1/(300*100*10⁻¹²) = 3.33*10⁷ ohm; I_rms = 230/3.33*10⁷ = 6.9*10⁻⁶ A = 6.9 microA. (b) Yes, by continuity the displacement current between the plates equals the conduction current. (c) Between the plates B at radius r < R: B0 = mu0 I0 r/(2 pi R²). I0 = sqrt(2)*6.9 microA = 9.76 microA. With r = 0.03, R = 0.06: B0 = (4 pi*10⁻⁷)(9.76*10⁻⁶)(0.03)/(2 pi (0.06)²) = 1.63*10⁻¹¹ T.

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