Question

Suppose that a parallel-plate capacitor has circular plates with radius R = 65.0 mm and a plate separation of 4.6 mm. Suppose also that a sinusoidal potential difference with a maximum value of 140 V and a frequency of 120 Hz is applied across the plates; that is

V=(140.0 V)sin((2.*π)*(120 Hz * t)).

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

b)Find B(r = 32.5 mm).

c)Find B(r = 130.0 mm).

d)Find B(r = 195.0 mm).

Answer 1

Here radius of the circular plates is R = 65 mm = 65*10^-3m

seperation between the plates is d = 4.6 mm = 4.6*10^-3m

frequency of voltage f = 120 Hz

Then angular frequency w = 2*pi*f = 2*pi*120 rad/sec

We know that magnetic field between the plates when r < = R is

Bin=(μ0*ε0*r/2)dE/dt

E = V/d

Then Bin=(μ0*ε0*r/2d)dV/dt

Here V = ( Vmax) sinwt

dV/dt = ( Vmax)wcoswt

dV/dt = (140 V) cos[2*pi(120 Hz)t]*2*pi(120Hz)

μ0= 4*pi*10^-7H/m

ε0= 8.85*10^-12C^2/Nm^2

Here Vmax = 140 V

This grows until r = R = 65 mm = 0.065 m

Then Bmax =(μ0*ε0*Rw/2d)Vmax

                  = [(4*pi*10^-7H/m) * (8.85*10^-12C^2/Nm^2)* 0.065 m *(2*pi*120 rad/sec )/2* 4.6 *10^-3m]*140Volts

                   = 1.755 *10^-16T