Induced voltage from current slope: A 120 mH coil carries a current changing at 150 mA/s. What magnitude of voltage is induced across the coil?

Introduction / Context:
Inductor voltage magnitude is proportional to the rate of change of current. Even relatively small slopes (in A/s) across moderate inductances generate measurable millivolt levels, which matter in precision analog design and sensor front-ends.

Given Data / Assumptions:

• L = 120 mH = 0.12 H.
• di/dt = 150 mA/s = 0.15 A/s.
• We seek |v| = L * |di/dt|; polarity depends on Lenz's law and is not requested.

Concept / Approach:
Use v = L * di/dt with consistent SI units. Multiply henries by amperes per second to get volts directly.

Step-by-Step Solution:

Verification / Alternative check:
Order-of-magnitude: 0.1 H * 0.1 A/s = 0.01 V (10 mV). Our numbers are slightly larger, giving 18 mV, which is reasonable.

Why Other Options Are Wrong:

• 180 mV: Off by a factor of 10.
• 1.8 mV and 1.25 mV: Too small; likely from arithmetic or prefix errors.

Common Pitfalls:

• Mistaking mA/s for A/s or mH for H, causing 10× or 100× errors.

Final Answer:
18 mV