A 10-cm-diameter parallel-plate capacitor has a 1.0 mm spacing. The electric field between the plates is increasing at the rate 1.5*106 V/ms. What is the magnetic field strength on the axis.

The magnetic field β = 3.1136 x 10 ⁻¹⁴ T

Explanation:

The miss information of r = 6.7 cm

The magnetic field can be find using the formula

β = μ0 ε0 (R² / 2r) dE/dt

Knowing the constant as μ0 and ε0

μ0 = 4π * 10⁻⁷ Tm/A

ε0 = 8.854 * 10^-12 C²/N∙m²

R = 0.050 m

r = 0.067 m

dE/dt = 1.5*10⁵ V/m∙s

β = 4π * 10⁻⁷ Tm/A * 8.854 * 10⁻¹² C²/N∙m² * (0.050²m / 2 * 0.067 m) * 1.5*10⁵ V/m∙s

β = 3.1136 x 10 ⁻¹⁴ T