Class 12 Physics
CBSE and UP Board
Electromagnetic Waves
Chapter-8 Exercise 8.2
NCERT Solutions for Class 12 Physics Chapter 8 Question 2
A parallel plate capacitor (Fig. 8.7) made of circular plates each of radius R = 6.0 cm has a capacitance C=100pF.The capacitor is connected to a 230 V ac supply with a (angular) frequency of to a 230 V ac supply with a (angular) frequency of 300 rad s⁻¹. (a) What is the rms value of the conduction current? (b) Is the conduction current equal to the displacement current? (c) Determine the amplitude of B at a point 3.0 cm from the axis between the plates.
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Radius of each circular plate, R = 6.0 cm = 0.06 m
Capacitance of a parallel plate capacitor, C = 100 pF = 100 x 10-12 F Supply voltage, V = 230 V
Angular frequency, ω = 300 rad s–1
Ans (a).
Rms value of conduction current, I = v/ xc
Where, Xc = Capacitive reactance =1/ωC
Therefore ,I = V x ωC = 230 x 300 x 100 x 10–12 = 6.9 x 10-6 A = 6.9 μA
Hence, the rms value of conduction current is 6.9μA.
Ans (b).
Yes, conduction current is equal to displacement current.
Ans (c).
Magnetic field is given as: B = μ0rI0/2πR2
Where,
μ0 = Free space permeability = 4π x 10-7 NA–2
Io = Maximum value of current = √2I
r = Distance between the plates from the axis = 3.0 cm = 0.03 m
Therefore, B = (4π x 10-7 x 0.03 x √2 x 6.9 x 10-6) /(2π x 0.062)
=1.63 x 10-11 T
Hence, the magnetic field at that point is 1.63 x 10-11 T.