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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 `c`is `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 isA. (a) clockwiseB. (b) anticlockwiseC. ( c) always changes with timeD. (d) cannot be calculated |
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Answer» Correct Answer - B (b) The large circiut is a with a time constant of `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 will act to oppose the decreases in flux from the large loop. Thus, the induced current flows counterclockwise . |
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