Suppose that a parallel-plate capacitor has circular plates with radius R 35.0 m

Question

Suppose that a parallel-plate capacitor has circular plates with radius R 35.0 mm and a plate separation of 4.9 mm. Suppose also that a sinusoidal potential difference with a maximum value of 340 V and a frequency of 60 Hz is applied across the plates; that is V= (340 V) sin [2 ? (60 Hz) t] Find Bmax(R), the maximum value of the induced magnetic field that occurs at r = R. Submit Answer Tries o/100 Find B(r17.5 mm) Submit Answer Tries 0/100 Find B(r = 70.0 mm) Submit Answer Tries 0/100 Find B(r 105.0 mm) Submit Answer Tries 0/100

Explanation / Answer

Capacitance of the parallel plate capacitor, C = A*epsilon/d

= pi*(35*10^-3)^2*8.854*10^-12/(4.9*10^-3)

= 6.95*10^-12 F

V = 340*sin(2*pi*60*t)

dV/dt = 340*cos(2*pi*60*t)*(2*pi*60)

= 128177*cos(2*pi*60*t)

dV_max/dt = 128177 V/s

Q = C*V

displacemnt current, Id = dQ/dt

= C*dV/dt

= 6.95*10^-12*128177

= 8.91*10^-7 A

Bmax(R) = mue*Id/(2*pi*R)

= 4*pi*10^-7*8.91*10^-7/(2*pi*35*10^-3)

= 5.10*10^-12 T <<<<<<<<---------Answer

B(r = 17.5 mm) = mue*Id*r/(2*pi*R^2)

= 4*pi*10^-7*8.91*10^-7*17.5*10^-3/(2*pi*(35*10^-3)^2)

= 2.54*10^-12 T <<<<<<<<---------Answer

B(r = 70 mm) = mue*Id/(2*pi*r)

= 4*pi*10^-7*8.91*10^-7/(2*pi*70*10^-3)

= 2.54*10^-12 T <<<<<<<<---------Answer

B(r = 105 mm) = mue*Id/(2*pi*r)

= 4*pi*10^-7*8.91*10^-7/(2*pi*105*10^-3)

= 1.70*10^-12 T <<<<<<<<---------Answer

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