Question

Group Problem 2: You are given the task to determine the inductance M of two inductors of unknown self-inductances L1 and L2. Recall that M = M12 = M21 and the magnetic flux B1,2 = M12/1 and B2,1 = M2112. To perform this task, you connect one of the inductors (say L1) in series with a battery of emf = 10 V and a resistor of resistance R1 = 212. These three components, as well as a switch S, form the circuit 1 in the figure below. The other inductor is connected in series with a resistor of resistance R2 = 2002 and a galvanometer G of negligible resistance, forming the circuit 2 in the figure. While the switch is still open, you bring the two circuits very close to each other so that the inductance coils overlap. After closing the switch, the current in circuit 1 increases until it eventually reaches its maximum value after a very long time. You observe that during this long time in which the current in circuit 1 was increasing, a total charge of 500 4C passed through the galvanometer in circuit 2. Yenibos Circuit 1 Circuit 2 (a) First, let us set up the problem. Using the definitions of self-inductance and mutual inductance. write the Kirchhoff's equations for both circuits. Note that they will be two coupled differential equations. Don't worry, we will not have to solve them. (b) Without solving these equations, what is the value of the currents in circuits 1 and 2 immediately after the switch S is closed? What about their values a very long time after the switch S has been closed? (c) To find the mutual inductance, we can focus only on the Kirchhoff's equation of circuit 2. Recall that the total charge Q that passed through the galvanometer is given just by the integral of the current in circuit 2 between the initial and final times, Q=S I2dt. Using the fact that Q1 = 500 C, the initial and final values of the currents in circuits 1 and 2 that you found in the previous item, and the value of R2, find the mutual inductance between the coils.