In the circuit shown in the following figure(Fiaure 1) the capacitor has capacit
ID: 1440093 • Letter: I
Question
In the circuit shown in the following figure(Fiaure 1) the capacitor has capacitance 15 muF and is initially uncharged. The resistor R_0 has resistance 12 ohm. An emf of 92.0 V is added in series with the capacitor and the resistor. The emf is placed between the capacitor and the switch, with the positive terminal of the emf adjacent to the capacitor. The small circuit is not connected in any way to the large one. The wire of the small circuit has a resistance of 1.4 ohm/m and contains 26 loops. The large circuit is a rectangle 2.0 m by 4.0 m, while the small one has dimensions a = 11.0 cm and b = 19.0 cm. The distance c is 6.0 cm. (The figure is not drawn to scale.) Both circuits are held stationary. Assume that only the wire nearest the small circuit produces an appreciable magnetic field through it. Part A The switch is closed at t = 0. When the current in the large circuit is 4.70 A, what is the magnitude of the induced current in the small circuit? Express your answer to two significant figures and include the appropriate units. Part B What is the direction of the induced current in the small circuit?Explanation / Answer
The current through the large circuit
:i = i0 (e^(-t/RC))
i = (V/R) (e^(-t/RC))
The rate of the current change:
di/dt = (V/R) (-1/RC) (e^(-t/RC))
di/dt = (V/CR^2) (e^(-t/RC))----------------------------------
we should find t
:i = (V/R) (e^(-t/RC))
4.70 = (92/12) e^(-t/((12)*(15e-6)))==>
t = 8.8077e-5 s----------------------------------
The magnetic field at a distance r:
B = u0 i/(2 pi r)
So
dB/dt = {u0/(2 pi r)}
{di/dt}dB/dt = {u0/(2 pi r)} {(V/CR^2) (e^(-t/RC))}
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The magnetic flux through the small circuit:
emf = dphi/dt emf = int{N A (dB/dt)}
emf = int{N (0.19 dr) {u0/(2 pi r)} {(V/CR^2) (e^(-t/RC))}}
emf = (0.19 N u0)/(2 pi) {(V/CR^2) (e^(-t/RC))} int{dr/r}
emf = (0.19 N u0)/(2 pi) {(V/CR^2) (e^(-t/RC))} {Ln(a+0.06) - Ln(a)}
emf = ((0.19)*(26)*(4*3.1416e-7))/(2*3.1416) * ((92/((15e-6)*(12)^2)) (e^(-(8.8077e-5)/((12)*(15e-6))))) {Ln(0.11+0.06) - Ln(0.11)}
emf = (0.000000988) * (42593) * (0.61304) * ((-1.772) - (-2.207275))
emf = 0.01123 V
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the resistance of the small circuit:
R' = (2) * (0.11+0.19) * (26) * (1.4) = 21.84 ohms
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the current:
i' = emf/R' = (0.01123)/(21.84) = 5.14 x 10^-4 A