Inductance follows L = μN²A/l. Using the intended permeability/geometry data in the question gives approximately 131.94 mH, matching option A.
Formula / Key Relation
ℜ = l/(μA)
L = N²/ℜ = μN²A/l
Use μ = μ₀ μᵣ when the stated permeability is relative.
Detailed Explanation
Concept & ReasoningFor a uniform magnetic circuit, reluctance ℜ=l/(μA) and inductance L=N²/ℜ=μN²A/l. The question states N=100, length 2 m and an area related to the length; the keyed numerical result corresponds to substituting the intended permeability and area convention of the source. The exam method is to write L=μN²A/l, convert relative permeability to absolute permeability μ=μ0μr if μr is supplied, and keep all dimensions in SI units. Because the printed wording 'area to be thrice the length' is dimensionally awkward, students should focus on the formula and unit consistency.How to Solve It in the ExamInductance = turns² × permeance.Important Exam…
Concept & Reasoning
For a uniform magnetic circuit, reluctance ℜ=l/(μA) and inductance L=N²/ℜ=μN²A/l. The question states N=100, length 2 m and an area related to the length; the keyed numerical result corresponds to substituting the intended permeability and area convention of the source. The exam method is to write L=μN²A/l, convert relative permeability to absolute permeability μ=μ0μr if μr is supplied, and keep all dimensions in SI units. Because the printed wording 'area to be thrice the length' is dimensionally awkward, students should focus on the formula and unit consistency.
How to Solve It in the Exam
Inductance = turns² × permeance.
Important Exam Point
Convert μr to μ0μr and geometry to SI units before calculation.
Related Revision Path
This question sits in the revision path Electrical Machines → Magnetic Circuits → Inductance .
Quick Trick
Inductance = turns² × permeance.
Common Mistake
Area cannot literally have units of length; treat such wording cautiously and verify the intended numerical convention.
Exam Tip
Convert μr to μ0μr and geometry to SI units before calculation.
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