The Total solution for NCERT class 6-12
A parallel plate capacitor (Fig.8.7) 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.
(a) What is the rms value of theconduction current?
(b) Is the conduction current equal to thedisplacement current?
(c) Determine the amplitude of B at a point 3.0 cm from the axisbetween the plates.
Radius of each circular plate, R = 6.0 cm = 0.06 m
Capacitanceof a parallel plate capacitor, C =100 pF = 100 × 10−12 F
Supplyvoltage, V = 230 V
Angularfrequency, ω = 300 rad s−1
Where,
XC =Capacitive reactance
∴ I = V × ωC
= 230 ×300 × 100 × 10−12
= 6.9 × 10−6 A
= 6.9 μA
Hence,the rms value of conduction current is 6.9 μA.
(b) Yes, conduction current is equal todisplacement current.
(c) Magnetic field is given as:
B
μ0 = Free spacepermeability
I0 = Maximum value ofcurrent =r =Distance between the plates from the axis = 3.0 cm = 0.03 m
= 1.63 × 10−11 T
Hence,the magnetic field at that point is 1.63 × 10−11 T.