Question

When current in the coil of 15 loops drops form 3.5A to 2.0A in 0.50 seconds, an average ε of 12mV is induced in that coil. What is the self-inductance in the coil? (4mH)

Answer 1

The self-inductance of a coil can be calculated using the formula L = - (ε / ΔI), where L is the self-inductance, ε is the induced emf, and ΔI is the change in current. In this case, the average induced emf ε is 12 mV and the change in current ΔI is 1.5 A. Plugging in these values, we find that the self-inductance of the coil is 8 mH.

Step-by-step explanation:

The self-inductance of a coil can be calculated using the formula L = - (ε / ΔI), where L is the self-inductance, ε is the induced emf, and ΔI is the change in current. In this case, the average induced emf ε is given as 12 mV, and the change in current ΔI is 3.5 A - 2.0 A = 1.5 A. Plugging in these values, we get L = - (12 mV / 1.5 A) = - 8 mV/A. However, self-inductance is always positive, so the magnitude of L is 8 mV/A. To convert this to the given unit of mH (millihenries), we divide by 1000 to get 8 mH.