Question

Suppose that a parallel-plate capacitor has circular plates with radius R = 25.0 mm and a plate separation of 4.8 mm. Suppose also that a sinusoidal potential difference with a maximum value of 180 V and a frequency of 60 Hz is applied across the plates; that is

V=(180.0 V)sin((2.*π)*(60 Hz * t)).

Find Bmax(R), the maximum value of the induced magnetic field that occurs at r = R.

Find B(r = 12.5 mm).

Find B(r = 50.0 mm).

Find B(r = 75.0 mm).

Answer 1

The maximum magnetic field will be given by:

B(max) = 0 0 r/2 x dE/dt =  0 0 r/2d x dV/dt since, E = V/d

B(max) =  0 0 r/2d x d((180.0 V)sin((2.*π)*(60 Hz * t))./dt

B(max) =  0 0 r x (180 V) (2 pi 60)/ 2d cos(2 pi 60 t)

for max B, cos term should be 1

B(max) at r = R

B(max) = 4 pi x 10^-7 x 8.85 x 10^-12 x 25 x 10^-3 x 180 x 2 pi 60 / 2 x 4.8 x 10^-3 = 1.96 x 10^-12 T

Hence B(max) at r = R is

B(max) = 1.96 x 10^-12 T

at ( r = 12.5 mm)

B(max) = 4 pi x 10^-7 x 8.85 x 10^-12 x 12.5 x 10^-3 x 180 x 2 pi 60 / 2 x 4.8 x 10^-3 = 0.98 x 10^-12 T

B(max) = 9.8 x 10^-13 T

at r = 50 mm

B(max) = 4 pi x 10^-7 x 8.85 x 10^-12 x 50 x 10^-3 x 180 x 2 pi 60 / 2 x 4.8 x 10^-3 = 3.92 x 10^-12 T

B(max) = 3.92 x 10^-12 T

at r = 75 mm

B(max) = 4 pi x 10^-7 x 8.85 x 10^-12 x 75 x 10^-3 x 180 x 2 pi 60 / 2 x 4.8 x 10^-3 = 5.88 x 10^-12 T

B(max) = 5.88 x 10^-12 T