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(8c32p7) Suppose that a parallel-plate capacitor has circular plates with radius

ID: 1298306 • Letter: #

Question

(8c32p7) Suppose that a parallel-plate capacitor has circular plates with radius R = 70.0 mm and a plate separation of 4.3 mm. Suppose also that a sinusoidal potential difference with a maximum value of 260 V and a frequency of 60 Hz is applied across the plates; that is

V=(260.0 V)sin((2.*?)*(60 Hz * t)).

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

b) Find B(r = 35.0 mm).

c) Find B(r = 140.0 mm).

d) Find B(r = 210.0 mm).

I've spent quite a long time on this, help is much appreciated!

Explanation / Answer

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Suppose that a parallel-plate capacitor has circular plates with radius R = 35.0 mm and a plate separation of 4.1 mm. Suppose also that a sinusoidal potential difference with a maximum value of 160 V and a frequency of 60 Hz is applied across the plates; that is
V=(160.0 V)sin((2.*p)*(60 Hz * t)).


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

Find B(r = 17.5 mm).

Find B(r = 70.0 mm).

Find B(r = 105.0 mm).

Answer

Here radius of the circular plates is R = 35 mm = 35*10-3m

seperation between the plates is d = 4.1 mm = 4.1*10-3m

frequency of voltage f = 60 Hz

Then angular frequency ? = 2?f = 2?*60 rad/sec

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

      Bin=(?0?0r/2)dE/dt

E = V/d

Then Bin=(?0?0r/2d)dV/dt

Here V = ( Vmax) sin?t

dV/dt = ( Vmax)?cos?t

dV/dt = (160 V) cos[2?(60 Hz)t]*2?(60Hz)

?0= 4?*10-7H/m

?0= 8.85*10-12C2/Nm2

Here Vmax = 160 V

This grows until r = R = 35 mm = 0.035 m

Then Bmax =(?0?0R?/2d)Vmax

                  = [(4?*10-7H/m) * (8.85*10-12C2/Nm2)* 0.035 m *(2?*60 rad/sec )/2* 4.1 *10-3m]*160Volts

                   = 2.86 *10-12T