Final answer:
The potential difference across an 11.0 mH inductor when the current drops from 130 mA to 60 mA in 12.0 µs is approximately -63.8 V. The negative sign indicates that the induced emf opposes the change in current.
Step-by-step explanation:
The potential difference across an inductor due to a change in current is given by Faraday's Law of electromagnetic induction. The formula used to calculate this is emf = -L (dI/dt), where emf is the electromotive force or potential difference in volts, L is the inductance in henrys (H), dI is the change in current in amperes (A), and dt is the change in time in seconds (s).
In this problem, we have an inductor with an inductance L of 11.0 mH (which is 11.0 x 10-3 H), a change in current dI of 70 mA (which is 70 x 10-3 A, since the current drops from 130 mA to 60 mA), and a change in time dt of 12.0 µs (which is 12.0 x 10-6 s).
Plugging these values into Faraday's Law, we get:
emf = -11.0 x 10-3 H * (70 x 10-3 A)/(12.0 x 10-6 s)
emf = -(11.0 x 70)/(12.0) V
emf ≈ -63.8 V
The negative sign indicates that the induced potential difference acts in a direction to oppose the change in current, in accordance with Lenz's Law.