An LC circuit has zero resistance, no battery, a capacitor C= 30microFarads, and

Question

An LC circuit has zero resistance, no battery, a capacitor C= 30microFarads, and an inductor L= 0.7 H connected in a simple loop. The capacitor and inductor store a combined total energy of U = 44 microJoules.
(a) If all the circuit's energy is stored in the capacitor at time t = 0, when is the soonest time that all the energy will be stored in the inductor?
     7.2 ms
(b) How much charge is stored on one side of the capacitor at t = 0?
     51.4 microCoulombs.
(c) How much current flows through the inductor at the time you found in (a)?
     11.2 mA
(d) If the inductor is a cylindrical solenoid of length l=4centimeters and radius r=2.8centimeters, find the number of coils in the solenoid.
     3010 coils
(e) What is the maximum magnetic field inside the inductor?
     1.06 mT
(f) If the maximum electric field in the capacitor is E = 1.2

Explanation / Answer

a) time = period/4 = (2*pi/w)/4 = (2*pi*sqrT(0.7*30.0E-6))/4= 7.2 ms

b) U= 1/2 Q^2/C

Q = sqrt( 2 U C) = sqrt(2*44.0E-6*30.0E-6) = 51.4 uC

c) E = 1/2 L I^2

I = sqrt( 2 E/L) = sqrt(2*44.0E-6/0.7) = 11.2 mA

d) L = u N^2 A/l

0.7 = 4*pi*10^(-7)*N^2*pi*2.8E-2^2/4.0E-2

N = 3010

e) B = u0 N I/L = 4*pi*10^(-7)*3010*11.2E-3/4.0E-2 = 1.06E-3 = 1.06 m T

f) Q = C V

E = V/d

E = Q/( C d)

d = Q/C E = 51.4E-6/(30.0E-6*1.2E3) = 1.43 mm

g) C = k e0 A/d

A = C d/k e0 = 30.0E-6*1.43E-3/(100000*8.85E-12) = 484 cm^2

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