Question

A parallel-plate capacitor with circular plates is charged by a constant current III. The radius aaa of the plates is much larger than the distance ddd between them, so fringing effects are negligible. Calculate B(r)B(r)B(r), the magnitude of the magnetic field inside the capacitor as a function of distance from the axis joining the center points of the circular plates.

Answer 1

Answer:

Please refer to the explanation below.

Explanation:

Let

Radius of parallel plate capacitor = r

Distance between parallel plates of the capacitor = R

Area of parallel plate capacitor = Ar

Area of Distance between parallel plates of the capacitor = AR

Constant Current = I₀

Permeability of free space = μ₀

Induced magnetic field = B

From the Ampere's Maxwell law the magnetic field is given by

∫B.ds = μ₀I₀

∫B.ds = μ₀I₀Ar/AR

When the radius of the plates is much less than the distance between the parallel plates (r < R)

∫B.ds = μ₀I₀πr²/πR²

∫B.ds = μ₀I₀r²/R²

s = 2πr

B.2πr = μ₀I₀r²/R²

B = μ₀I₀r²/2πrR²

B = μ₀I₀r/2πR²

When the radius of the plates is much larger than the distance between the parallel plates (r > R)

∫B.ds = μ₀I₀

B.2πr = μ₀I₀

B = μ₀I₀/2πr

Bonus:

The magnetic field will be maximum when the radius of the parallel plate capacitor is equal to he distance between the parallel plate capacitor that is r = R

B = μ₀I₀/2πR