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A parallel-plate air-filled capacitor is being charged as in the figure below. T

ID: 2030142 • Letter: A

Question

A parallel-plate air-filled capacitor is being charged as in the figure below. The circular plates have radius 6.00 cm, and at a particular instant the conduction current in the wires is 0.220 A 2 (a) What is the displacement current density jo in the air space between the plates? A/m2 (b) What is the rate at which the electric field between the plates is changing? V/m s (c) What is the induced magnetic field between the plates at a distance of 3.00 cm from the axis? (d) What is the induced magnetic field between the plates at 1.50 cm from the axis?

Explanation / Answer

Given,

r = 6 cm ; I = 0.220 A

a)We know that

J = I/A

J = 0.220/3.14 x 0.06^2 = 19.46 A/m^2

Hence, J = 19.46 A/m^2

b)Rate at which E field changes will be

dE/dt = J/e0

dE/dt = 19.46/(8.85 x 10^-12) = 2.2 x 10^12 V/m-s

Hence, dE/dt = 2.2 x 10^12 V/m-s

c)r = 3 cm = 0.03 m

The induced B field will be given by:

B = u0 i/2 pi x r/R^2

B = 4 pi x 10^-7 x 0.220/2 pi (0.03/0.06^2) = 3.67 x 10^-7 T

Hence, B = 3.67 x 10^-7 T

d)r = 1.5 cm = 0.015 m

using the same expression:

B = 4 pi x 10^-7 x 0.220/2 pi (0.015/0.06^2) = 1.83 x 10^-7 T

Hence, B = 1.83 x 10^-7 T