Eddy-Current Loss vs Frequency in Magnetic Cores How does eddy-current power loss in a homogeneous magnetic core vary with excitation frequency (keeping other factors like peak flux density and thickness comparable)?

Difficulty: Easy

Correct Answer: Proportional to (frequency)^2

Explanation:


Introduction / Context:
Eddy currents are circulating currents induced in conducting cores by time-varying magnetic flux. They cause I^2R losses and heating, reducing efficiency in transformers, motors, and inductors. Understanding their frequency dependence guides core material selection and lamination strategies.


Given Data / Assumptions:

  • Uniform sheet/lamination of thickness t and resistivity ρ.
  • Sinusoidal excitation; peak flux density Bmax considered comparable.
  • Classical eddy-current loss model applies.


Concept / Approach:

The classical expression for eddy-current loss density is Pe ∝ (Bmax^2) * (f^2) * (t^2) / ρ. Thus, with Bmax, t, and ρ fixed, eddy-current loss scales with the square of frequency. This is why thin laminations, high-resistivity ferrites, or powder cores (with distributed air gaps) are used at higher frequencies to suppress eddy currents.


Step-by-Step Solution:

Recall proportionality: Pe ∝ Bmax^2 * f^2 * t^2 / ρ.Hold Bmax, t, ρ constant → Pe ∝ f^2.Therefore, doubling frequency quadruples eddy-current loss.


Verification / Alternative check:

Core loss charts separate hysteresis (∝ f) and eddy-current (∝ f^2) components; measured total loss fits Steinmetz-like models combining both contributions.


Why Other Options Are Wrong:

  • Linear, cubic, or inverse dependences do not match the classical result.
  • Frequency independence contradicts Faraday’s law and induced emf scaling with f.


Common Pitfalls:

Confusing total core loss (which includes hysteresis and anomalous loss) with the eddy-current component alone; ignoring the role of lamination thickness.


Final Answer:

Proportional to (frequency)^2

More Questions from Materials and Components

Discussion & Comments

No comments yet. Be the first to comment!
Join Discussion