A capacitor with parallel circular plates of radius R = 1.20 cm is discharging v

Question

A capacitor with parallel circular plates of radius R = 1.20 cm is discharging via a current of 12.0 A. Consider a loop of radius R/3 that is centered on the central axis between the plates. (a) How much displacement current is encircled by the loop? The maximum induced magnetic field has a magnitude of 12.0 mT. At what radius (b) inside and (c) outside the capacitor gap is the magnitude of the induced magnetic field 3.00 mT?

Explanation / Answer

j(d) or current density = i(d)/A = i(c)/A = 12.5/pie*(0.0310)^2 = 4.14 x 10^3 Amps per meter squared now i(d) inside R/3 = j(d)*area = (4.14kA/m^2)*(pie*(0.0310^2/9)) = 1.39 Amps last time i said that the magnetic field is too small for the given current. but now i think the question is giving the maximum bfield "inside" or between the capacitor plates for any unknow current. so lets solve for that current: B=mu*I/2*pie*r therefore I = B *2*pie*r/mu =11.0mT*2*pie*(0.0310m)/mu = 1.705 x 10^3 amps of current is need to produce the desired maximum bfield. now bfield decrease as r increases or as it decreases we will have one r outside plates (r>R) and one inside (rR, B=mu*I/2*pie*r therefore r = mu*I/2*pie*B =mu*1.705kA/2*pie*(2.5mT) = 13.6 cm. for r

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