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In the circuit shown in Fig. the capacitor has capacitance C = 20 muF and is initially charged to 100 V with the polarity shown. The resistor R_(0) has resistance 10 Omega. At time t = 0, the switch is closed. The smaller circuit is not connected in any way to teh larger one. the wire of the smaller circuit has a resistance of 1.0 Omega m^(-1) and contains 25 loops. the larger circuit is a rectangle 2.0 m by 4.0 m, while the smaller one has dimensions a = 10.0 cm and b = 20.0 cm. the distance cis 5.0 cm. (The figure is not drawn to scale.) Both circuit are held stationary. Assume that only the wire nearest to teh smaller circuit produces. an appreciable magnetic field through it. The direction of current in the smaller circuit is |
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Answer» (a) CLOCKWISE `tau = RC = (10 Omega)(20 xx 10^(-6)F) = 200 mus`. Thus, the currnt as a function of time is `I = ((100 V)/(10 Omega))e^((-t)/(200mus))` At `t = 200 ms`, we obtain `I = (10 A)(e^(-1)) = 3.7 A`. Assuming that only the long wire nearest the small loop produces an appreciable magnetic flux through the small loop, `Phi_(B) = int_( c)^(c + a)(mu_(0)ib)/(2pi)dr = (mu_(0)ib)/(2pi)1n(1 + (a)/(c ))` ltbr. So the emf induced in the small loop at `t = 200 ms` `epsilon = -(dPhi)/(dt) = -(mu_(0)ib)/(2pi)1n(1 + (a)/(c ))(DI)/(dt)` `= -(((4pi xx 10^(-7)(Wb))/(A xx m^(2)))(0.200 m))/(2pi) xx 1n (3.0)(-(3.7 A)/(200 xx 10^(-6) s))` Thus, the induced current in the small loop is `i = (epsilon)/(R ) = (0.81 MV)/(25(0.600 m)(1.0 Omega//m)) = 54 muA`. ltbr. Initially current in large loop is maximum and afterwards decreases. Hence flux through the SMALLER loop decreases with time. The induced current willact to oppose the decreases in flux from the large loop. Thus, the induced current flows counterclockwise . |
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