The displacement current in the middle of a capacitor as it charges is 13A. The

Question

The displacement current in the middle of a capacitor as it charges is 13A. The voltage across the charging capacitor is increasing at a rate of 1.10 million V/s. The cross-sectional area of the capacitor is 0.200cm^2. What is the rate of change of the electric field through the center of the capacitor (dE/dt)? What's the capacitance of the capacitor? At what distance from a 10W point source of electromagnetic waves (light) is the magnetic field amplitude 1.0 mu T? Choose two of the following applications of Faraday's Law: car sensors at stoplights, credit card readers, microphones, or metal detectors. Explain how these technologies apply the law and, at a basic physics level, how they work.

Explanation / Answer

givens:
Psource = 10 W
B0 = 1.0T
and we're trying to determine the distance from the point source, r

How can we relate the magnetic field amplitude and the distance from the point source?
neither of our basic field formulas helps us, but we do know a series of equations relating the Electric field amplitude and electromagnetic wave intensity, and the amplitudes of the Electric field and the Magnetic field are related by the speed of light, we can arrange the following formulas to solve for r

Intensity relations~
I= Psource1/(4r2) = c0E021/2
Magnetic & Electric field relations~
B0c = E0
so E02 = (B0c)2
I= c0(B0c)21/2=c30(B0)21/2
since I also equals Psource1/(4r2), it can be simplified that
Psource1/(4r2) = c30(B0)21/2
now, 0 = 1/(0c2) = 1/[(410-7 H/m)c2]
so Psource1/(4r2) = c31/[(410-7 H/m)c2](B0)21/2 simplifies to
Psource[210-7 H/m] / (r2) = c(B0)2 which simplifies to
r2=Psource(210-7 H) 1/[cB02 m]
leaving r = (Psource(210-7 H) 1/[cB02 m])
now, plugging in for all of the values for which we've arranged so nicely, we get
r = {(10 W)(210-7 H)1/[(3108 m/s)(1*10-12)T)2 m] }


{(10)(210-7)1/[310-4] = {6.66*10-3} = 1/6 = 8.16*10-2 m

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