Question

Suppose that a parallel-plate capacitor has circular plates with radius R = 60.0 mm and a plate separation of 4.6 mm. Suppose also that a sinusoidal potential difference with a maximum value of 360 V and a frequency of 120 Hz is applied across the plates; that is

V=(360.0 V)sin((2.*π)*(120 Hz * t)).

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

Find B(r = 30.0 mm).

Find B(r = 120.0 mm).

Find B(r = 180.0 mm).

Answer 1

R =60 mm , d =4.6 mm,

V = (360V) sin((2.*π)*(120 Hz * t)).

From Ampheres law

B.dl = o id

id =eo (dE/dt) =eo AdE/dt

E = V/d

V=(360.0 V)sin((2.*π)*(120 Hz * t)).

dE/dt = (360)(240π)cos((2.*π)*(120 Hz * t))/d

id =eo (πR²) (360)(240π)cos((2.*π)*(120 Hz * t))/d

B.dl = B(2πr) = oeo (πR²) (360)(240π)cos((2.*π)*(120 Hz * t))/d

oeo =1/c^2

for maximum B value

Bmax = 43200 (πR²) /rdc²

For r =R

Bmax = 43200 (πR) /dc²

Bmax = (43200*3.14*0.06)/(0.0046)(3*10⁸)²

Bmax = 1.966x10^-11 T

For r =30 mm

Bmax = 43200 (πR²) /rdc²

= 43200(3.14*60*60)/(30*4.6)(3*10⁸)²

Bmax = 3.93x10^-11 T

For r = 120 mm

Bmax = 43200 (πR²) /rdc²

= 43200(3.14*60*60)/(120*4.6)(3*10⁸)²

Bmax = 0.983x10^-11 T

For r =180 mm

Bmax = 43200 (πR²) /rdc²

= 43200(3.14*60*60)/(180*4.6)(3*10⁸)²

Bmax = 0.655x10^-11 T