The current in a coil with a self-inductance of 2.4 mH increases from 0 to 2.66 A in a tenth of a second.What is the induced emf in the coil?
A)64 mV
B)6.4 mV
C)1 700 mV
D)5.1 V
E)130 mV

Question

Quizplus · Accepted Answer

The induced emf (ε) in a coil can be calculated using the formula ε = -L(ΔI/Δt), where L is the self-inductance, ΔI is the change in current, and Δt is the change in time. Here, L = 2.4 mH = 0.0024 H, ΔI = 2.66 A, and Δt = 0.1 s. Plugging these values in, we get ε = -0.0024 H * (2.66 A / 0.1 s) = -0.06384 V = -63.84 mV. The negative sign indicates the direction of the induced emf, which is not needed for the magnitude answer, so the magnitude is 63.84 mV, which is approximately 64 mV.

The Current in a Coil with a Self-Inductance of 2