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
H = mc∆T
where, H=heat energy
c = SHC
H = 2*921*(40-20)
H = 36840J
H = 36.84KJ
The energy transferred to it is 36.84KJ
The penny will reach terminal velocity at 50 ft. Then it will travel 25 mph till it reaches ground.
The most common metals used for permanent magnets are iron, nickel, cobalt and some alloys of rare earth metals