Final answer:
To find the potential difference across an 11.0 mH inductor with a decrease in current from 130 mA to 60 mA over 12.0 μs, the induced emf formula ε = -L(ΔI/Δt) is used, yielding a potential difference of 64.1667 V.
Step-by-step explanation:
The potential difference across an inductor can be calculated using the formula for induced emf in an inductor, which is ε = -L(ΔI/Δt), where ε is the induced emf, L is the inductance, ΔI is the change in current, and Δt is the change in time. In this case, the inductance L is 11.0 mH (or 0.011 H), the change in current ΔI is 130 mA - 60 mA (or 0.130 A - 0.060 A), and the change in time Δt is 12.0 μs (or 12.0 x 10^-6 s).
First, we calculate ΔI:
ΔI = 0.130 A - 0.060 A = 0.070 A
Now, we substitute the given values into the formula:
ε = -0.011 H * (0.070 A / 12.0 x 10^-6 s)
ε = -0.011 H * (5833.33 A/s)
ε = -64.1667 V
The negative sign indicates that the induced emf opposes the decrease in current, according to Lenz's Law. Therefore, the potential difference across the inductor is 64.1667 V.