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3 years ago

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A parallel-plate capacitor with circular plates is charged by a constant current III. The radius aaa of the plates is much large

r 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.

1 answer:

shutvik [7]3 years ago

4 0

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

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