Compute reluctance: A magnetic material has length l = 0.07 m, cross-sectional area A = 0.014 m^2, and permeability µ = 4,500 µWb/(At·m). Find the reluctance ℜ in At/Wb.

Difficulty: Medium

Correct Answer: 1111 At/Wb

Explanation:


Introduction / Context:
Reluctance determines how much magnetomotive force is required to establish a given magnetic flux in a magnetic path. It mirrors electrical resistance in Ohm’s law, with the magnetic counterpart given by Hopkinson’s law: MMF = ℜ * phi.


Given Data / Assumptions:

  • Length, l = 0.07 m.
  • Area, A = 0.014 m^2.
  • Permeability, µ = 4,500 µWb/(At·m) = 4,500 * 10^-6 Wb/(At·m) = 0.0045 Wb/(At·m).
  • Uniform cross-section and material properties; linear operation (no saturation considered).


Concept / Approach:

The reluctance formula is ℜ = l / (µ * A), with units At/Wb. Substitute SI-consistent values, being careful with the micro (10^-6) prefix in permeability.


Step-by-Step Solution:

Convert µ: µ = 4,500 µWb/(At·m) = 0.0045 Wb/(At·m).Compute µA: µ * A = 0.0045 * 0.014 = 6.3 * 10^-5 Wb/At.Compute ℜ: ℜ = l / (µA) = 0.07 / (6.3 * 10^-5) ≈ 1.111 * 10^3 At/Wb.Therefore, ℜ ≈ 1111 At/Wb.


Verification / Alternative check:

Dimensional check: m / (Wb/(At·m) * m^2) = m / (Wb/At * m) = At/Wb, correct unit for reluctance.


Why Other Options Are Wrong:

11 and 111 At/Wb underestimate by factors of 100 and 10. 1 At/Wb is far too small. 7 At/Wb has no basis in the calculation.


Common Pitfalls:

Failing to convert µ from micro-units, or using µ_r instead of µ (absolute) in the formula. Always ensure SI consistency.


Final Answer:

1111 At/Wb

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