Final answer:
In a circuit with two coils with a mutual inductance of 5.00 mH, when a 2.00 A current is switched off in one coil over 30.0 ms, the magnitude of the induced voltage in the other coil would be 333 mV.
Step-by-step explanation:
The original question seems to have typos and missing context, however, it appears to ask about induced current in a secondary circuit when there is a change in current in the primary circuit. To answer a similar valid question:
Given that two coils have a mutual inductance of 5.00 mH (millihenrys), we can find the voltage induced in one coil when the current in the other coil is changed. According to Faraday's Law of Electromagnetic Induction, the induced voltage (E) in a coil is equal to the negative rate of change of magnetic flux, which depends on the mutual inductance (M) and the rate of change of current (ΔI/Δt). The formula is E = -M (ΔI/Δt).
If a 2.00 A current in one coil is switched off over a period of 30.0 ms, then the induced voltage in the other coil can be calculated using the given mutual inductance:
E = -5.00 mH * (0 A - 2.00 A) / 30.0 ms
E = -5.00 mH * (-2.00 A) / 0.030 s
E = (5.00 x 10^-3 H) * (2 A / 0.030 s)
E = 0.333 V
So the magnitude of the induced voltage would be 333 mV (millivolts).