Answer:
Given that
capacitor being charged by a current i c has a displacementcurrent equal to i c between the plates∴
displacement current iD =i c=0.280 Aradius of the circular plate r = 4 cm=0.04 m
( A ) . displacement current density j D = iD / ( π r 2 )=0.28 / ( 3.14 * 0.04 2 )=55.73 A / m 2
( B ) . displacement current density j
D = ε( dE / dt )the rate at which the electric field between the plates is changing(
dE / dt ) = jD/ εdE/ dt ) = 55.73/ 8.85 * 10 -12=6.3*10 12 N / C - s( C ) . the induced magnetic field between theplates B = ( μ * r / 2π R 2) * i c ----( 1 )whereR= 2 cm=0.02 mr= 4 cm=0.04 mμ=permeability of free space=4π* 10 -7 H( D )substitute R = 1 cm = 0.01 m inequation ( 1 ),wegetanswer