Compute magnetomotive force (MMF) from turns and current For a coil with N = 150 turns carrying current I = 2 A, what is the magnetomotive force F = N * I (in ampere-turns)?

Introduction / Context:Magnetomotive force (MMF) is the magnetic analog of electromotive force in circuits. It drives magnetic flux through a magnetic path and is directly proportional to the number of turns and the current through those turns. Calculating MMF is a foundational step in designing inductors, solenoids, and transformers.

Given Data / Assumptions:

• Number of turns N = 150 turns.
• Current I = 2 A.
• Use F = N * I with F in ampere-turns (At).

Concept / Approach:The greater the product of turns and current, the stronger the magnetizing force. While material properties and geometry affect resulting flux, MMF itself is purely a function of windings and current, independent of core type in the basic calculation.

Step-by-Step Solution:Write the formula: F = N * I.Substitute: F = 150 * 2 = 300 At.Therefore, MMF = 300 ampere-turns.

Verification / Alternative check:Units check: turns × amperes → ampere-turns; no conversions needed.

Why Other Options Are Wrong:13.33 mAt: off by orders of magnitude; milli-At inappropriate here.75 At: corresponds to 150 * 0.5 A, not given.152 At: not any simple combination of the provided values.

Common Pitfalls:Confusing MMF with magnetic field intensity H; H also depends on magnetic path length: H = F / l. Here only MMF is requested.

Final Answer:300 At