Question

Suppose that a parallel-plate capacitor has circular plates with radius R = 34 mm and a plate separation of 6.9 mm. Suppose also that a sinusoidal potential difference with a maximum value of 120 V and a frequency of 51 Hz is applied across the plates; that is, V = (120 V) sin[2π(51 Hz)t]. Find Bmax(R), the maximum value of the induced magnetic field that occurs at r = R.

Answer 1

given that

radius of the circular plates is R = r = 34 mm = 0.034 m

seperation between the plates is d = 6.9 mm = 0.0069 m

frequency of voltage f = 51 Hz

Then angular frequency w = 2*pi*f = 2*3.14*51 rad/sec

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

Bin = (u0*0*r/2)*dE/dt

E = V/d

Bin = (u0*0*r/2d)*dV/dt

Here V = ( Vmax) sinw*t

dV/dt = ( Vmax)*w*cosw*t

dV/dt = (120 ) *cos[2*3.14*51*t]*2*3.14*51

u0 = 4*pi*10^(-7) H/m

0 = 8.854*10^(-12) C^2/Nm^2

Bmax =(u0*0*R*ω/2d)*Vmax

Bmax = ( 4*3.14*10^(-7)*8.85*10^(-12)*0.034*2*3.14*51 /2*0.0069 )*120

Bmax = 10525391.30*10^(-19)

Bmax = 1.05*10^(-12) T