Final answer:
The rate of change of the current in the smaller inductor is the same as that in the larger one, 1200A/s, because the voltage across both inductors in parallel is the same.
Step-by-step explanation:
The question is asking for the rate of change of current in a smaller inductor when two inductors, one with inductance L and the other with inductance 2L, are connected in parallel and the rate of change of the current in the larger inductor is known. In a parallel circuit, the voltage across both inductors must be the same. Since the induced emf (voltage) in an inductor is given by V = L(dI/dt), and both inductors have the same voltage across them, the rate of change of current can be determined using the ratio of their inductances.
Given that the larger inductor with 2L has a rate of change of 1200 A/s, the smaller inductor with L will have a rate of change of current that is half of that, because V is constant and L for the smaller inductor is half the size. Therefore, the rate of change of the current in the smaller inductor will also be 1200 A/s. The correct answer is B. 1200A/s.