Respuesta :
a) Displacement current density for the capacitor is 113.4 A/m²
b) Rate of change of the electric field is 1.48157826 V/m.s
c) Magnetic field between plates at r of 1.93 cm is 5.563×10⁻6 Tesla
d) Magnetic field between plates at r of 1.10 cm is 7.833 x 10-7 Tesla
Since we are given the radius of a parallel-plate capacitor which is 4.00 cm( 0.04 m) and the Conduction current = Ic = 0.570 A .we know the formula for the Displacement current density is
JD = Id/A, where Id is the conduction current and A is the area of the capacitor, and A= πr² = π(0.04)²= 0.005024 m². As we know conduction current has equal displacement between the capacitor plates so Id = Ic
JD = 0.570/0.005024 = 113.4 A/m²
For the second case, we know the rate of change of the electric field is:
dE/dt = JD/ε₀ , where JD is the displacement current density and ε₀ is the permittivity of free space ( 8.854×10⁻¹² C²/N.m²)
so, dE/dt = 113.4/8.854×10⁻¹² =1.48157826 V/m.s
for the third case, the formula for The induced magnetic field between the plates will be :
B = (μ₀/2)*JD*r, μ₀ is the permeability of free space(4π×10⁻⁷ T.m/A)and r is the distance from the axis.
B = (4π×10⁻⁷/2)*113.4 *0.01953
B = 5.563× 10⁻⁶ Tesla ,and for the last case, for the radius of 1.10 cm (0.011 m)
B = (μ₀/2)*JD*r
B = (4π×10⁻⁷/2)*113.4 *0.011
B = 7.833 x 10⁻⁷ Tesla
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