In the circuit shown in the figure below the capacitor has capacitance C = 30 mu
ID: 2300094 • Letter: I
Question
In the circuit shown in the figure below the capacitor has capacitance C = 30 mu F and is initially charged to 100 V with the polarity shown. The resistor R0 has resistance 10 ohm. At time t = 0 the switch is closed. The small circuit is not connected in any way to the large one. The wire of the small circuit has a resistance of 1.0 ohm/m and contains 27 loops. The large circuit is a rectangle 2.0 m by 4.0 m, while the small 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 circuits are held stationary. Assume that only the wire nearest the small circuit produces an appreciable magnetic field through it. Find the current in the large circuit 240 ps after S is closed. Find the current in the small circuit 240 ps after S is closed.Explanation / Answer
current in large circuit at t= 240 micro secs = 4.49 A (given in question)
now
small circuit
induced emf = N*dB/dt*A
current = E/R
area = 10*20*10^-4 = 0.02 square meter
resistance of small circuit = 2(l+b)* Resistance/meter = 60/100 * 1 = 0.6 ohms
B = mu_o * I / (2*pi*c) {since dimensions of large circuit are greater than smaller one}
= 4*pi* 10^-7 * 4.49 / (2*pi*0.05) = 4*4.49 micro tesla
now
E = N*B*A = 27*4*0.02*4.49 = 2.16*4.49 micro volts
current = 2.16*4.49 / 0.6 = 16.164 micro amps
so answer is 16.164 micro amps