Final answer:
Using the formula for induced emf in an inductor, the calculated induced emf when the current through a 7.50 mH inductor is switched off in 8.33 ms is 3.6 V. However, this value does not match the answer choices, suggesting a possible typo in those choices.
Step-by-step explanation:
When the 4.00 A current through a 7.50 mH inductor is switched off in 8.33 ms, the induced emf opposing the change can be calculated using Faraday's law of electromagnetic induction, which states that the magnitude of the induced emf (E) is equal to the negative of the rate of change of the magnetic flux. In the case of an inductor, this translates to E = -L * (ΔI/Δt), where L is the inductance, ΔI is the change in current, and Δt is the time interval over which the current changes.
For this problem, L = 7.50 mH (or 0.0075 H), ΔI = 4.00 A (since the current goes from 4.00 A to 0 A), and Δt = 8.33 ms (or 0.00833 s). The negative sign indicates the direction of the induced emf is opposite to the change in current, according to Lenz's Law.
Now, we can calculate the induced emf as E = -0.0075 H * (4.00 A / 0.00833 s) = -3.6 V. The negative sign simply indicates the direction of the induced voltage; the magnitude of the induced emf is 3.6 V. However, given the answer choices provided, it seems there might be a calculation or typo error since 3.6 V is not an option. If the intention was to give the answer in volts, the correct calculation methodology has been presented, but the students should re-examine the choices or the initial conditions provided for a potentially typo.