A parallel plate capacitor is made using two circular conducting plates that eac

Question

A parallel plate capacitor is made using two circular conducting plates that each has a radius of 55.2 cm. The two plates are separated by a distance of 5.2 mm. A generator supplying an alternating emf of V(t) = 15.0 V sin [2pi(7.25 MHz)t] is connected across the capacitor. Determine the amplitude of the magnetic field at a distance of 20.0 cm from the axis of the capacitor, between the capacitor plates. Give your answer in the form "a.bc x 10^(y) T". (Note: pi = 3.14.) Here are hints: Use Ampere's law and assume that the electric field between the plates does not depend on the distance from either capacitor plate but is time dependent. You will also need the displacement current density.

Explanation / Answer

From Amperes law
closed integra of B. dl = uo times of ( I+ Id)

Dispacement id = eeo dphi/dt

so In between the plates I +Id = eo d phi/dt

but EA = q/eo = phi

so

dphi/dt = 1/eo * dq/dt

B. closed dl = uo dq/dt

B* 2pi R = uo dq/dt

so magnetic field B = (uo/2piR) * dq/dt

bu for capacitor charge Q = CV = eo A/d

so
magnetic field B = (uo/2pir) * (eo pi r^2/d)

B = 15 * 7.5 e 6 * 2pi cos 2pi(7.25 e 6) t

B = 4.03 e -7 Tesla or 0.403 uT

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