Self-induced emf = 13.0 mV
DI/dt: 10.0 A/s
Current: 5.00 A
We just need to manipulate a few equations for inductance to solve this. We know that
self-induced emf = -L * (DI/dt)
Where L represents inductance. We know that inductance is:
L = ((N * ΦB)/I)
Where ΦB is magnetic flux, N is the number of turns in the coil, and I is the current. With this information, we can substitute the equation for inductance into the equation for self-induced EMF.
self-induced emf = -((N * ΦB)/I) * (DI/dt)
From the prompt, we have the values for self-induced emf, N, I, and DI/dt. This leaves the value for magnetic flux as the only value, which only requires some quick algebraic manipulation to solve!