MasteringPhysics: Honors credit Assignment 03 Google Chrome https://session.mast
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Question
MasteringPhysics: Honors credit Assignment 03 Google Chrome https://session.masteringphysics.com/myct/itemView?assignmentProblemID 728404048toffset-next signed in as quanyou jiang l Help Close Physics 2760, University Physics ll. Spring 2017 Resources T Honors Credit-Assignment 03 Problem 29 48 previous l 7 of 8 l nexi Problem 29.48 Part A In the circuit shown in the following figure(Figure 1)the The switch is closed at t 0. When the current in the large circuit is 3.40 A, what capacitor has capacitance 19 AuF and is initially is the magnitude of the induced current in the small circuit? uncharged. The resistor Ro has resistance 10 An emf of 92.0 V is added in series with the capacitor and Express your answer to two significant figures and include the appropriate the resistor The emf is placed between the capacitor units. 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 1.50/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 14.0 cm and 0.025 A b 24.0 cm. The distance c is 30 cm mhe figure is not drawn to scale Both circuits are held stationary Assume that only the wire nearest the small circuit Submit My Answers Give Up produces an appreciable magnetic field through it incorrect; Try Again; 4 attempts remaining Part B What is the direction of the induced current in the small circuit? counterclockwise clockwise Submit MAnswers Give Up Figure 1 of 1 Correct Continue https//session mastering physi 72840405&offset; nextExplanation / Answer
i(t) = (E/R) * e^-t/RC
Assuming that only the long wire nearest the small loop produces an appreciable magnetic flux through he small loop
phi_b = integral uo*i*b *dr/2pir from c to c +a = (uo*i*b/2pi )*ln(1+a/c)
Einduced = -dphi_b/dt
by chain rule
dphi_b/dt = (uo*b/2pi)*ln(1+a/c)*di/dt * di/t = -E/R^2C * e^-t/RC
Einduced = N|dphi_b/dt| = N*uo*b/2pi * ln(1+a/c) *i/RC
R of the small loop = 1.5 * 26 * 0.76 = 29.64 ohm
Einduced = (26 * 2 x 10^-7 * 0.24 )*ln(1+14/6) * 3.40/(10 * 19 x 10^-6) = 0.02689 V
induced current = Einduced/R = 9.07 x 10^-4 A = 0.00091 A