Question

A parallel-plate capacitor with circular plates of radius 40 mm is being discharged by a current of 6.0 A. At what radius (a) inside and (b) outside the capacitor gap is the magnitude of the induced magnetic field equal to 75% of its maximum value?(c) What is that maximum value?

Answer 1

Answer:

A) r = 0.03 m

B) r = 0.0533 m

C) B_max = 0.00003 T

Explanation:

Formula for magnetic field inside the capacitor when it is parallel to the length element is;

B_in = (μ_o•I•r/(2πR²)

Formula for maximum magnetic field is;

B_max = (μ_o•I/(2πR)

Formula for magnetic field outside the capacitor is; B_out = (μ_o•I/(2πr)

A) Magnetic field inside the capacitor is gotten from our first equation above;

B_in = (μ_o•I•r/R²)

Since we want to find the radius at which the magnitude of the induced magnetic field equal to 75% or 0.75 of its maximum value.

Thus;

B_in = 0.75B_max

(μ_o•I•r/(2πR²) = 0.75((μ_o•I/(2πR))

μ_o•I and 2πR will cancel out to give;

r/R = 0.75

r = 0.75R

We are given R = 40 mm = 0.04 m

r = 0.75 × 0.04

r = 0.03 m

B) magnetic field outside the capacitor is; B_out = (μ_o•I/(2πr)

Thus for the magnitude of the induced magnetic field equal to 75% or 0.75 of its maximum value:

B_out = 0.75B_max

(μ_o•I/(2πr) = 0.75((μ_o•I/(2πR))

μ_o•I and 2π will cancel out to give;

1/r = 0.75/R

r = R/0.75

r = 0.04/0.75

r = 0.0533 m

C) B_max = μ_o•I/(2πR)

μ_o is a constant known as vacuum of permeability with a value of 4π × 10^(-7) T.m/A

Thus;

B_max = (4π × 10^(-7) × 6)/(2π × 0.04)

B_max = 0.00003 T