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) sinw*t*

dV/dt = ( Vmax)wcosw*t*

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

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