Final answer:
The self-inductance of the coil is calculated using the formula L = E ∆t / ∆I with the induced emf of 0.012V, the change in current of 0.8 A, and the change in time of 0.52 s, resulting in an self-inductance of 7.8 mH.
Step-by-step explanation:
The student is asking about the self-inductance of a coil based on the rate of change of current and the induced electromotive force (emf). To find the self-inductance of the coil, the formula used is E = -L ∆I/∆t, where E is the induced emf, L is the self-inductance, ∆I is the change in current, and ∆t is the change in time. Given that the induced emf (E) is 12mV or 0.012V, the change in current (∆I) is 3.4 A - 2.6 A = 0.8 A, and the change in time (∆t) is 0.52 s, we can rearrange the formula to solve for L, resulting in L = E ∆t / ∆I. Plugging the values into this equation, we get L = (0.012 V) × (0.52 s) / (0.8 A). After calculating, we determine that the self-inductance of the coil is 0.0078 H or 7.8 mH.