An iron rod of area of cross-section 0.1m2 is subjected to a magnetising field of 1000 A/m. Calculate the magnetic permeability of the iron rod. [Magnetic susceptibility of iron = 59.9, magnetic permeability of vacuum = 4π x 10-7 S. I. unit]
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Solution
Given:- H= 1000 A/m, χ = 59.9, μ0 = 4π x 10-7 S.I. unit
To find:- Permeability (μ)
Formula:- μ = μ0 (1 + χ)
Calculation:- From formula,
μ = 4π x 10-7 (1 + 59.9)
= 4 x 3.142 x 10-7 x 60.9
= antilog [log(4) + log(3.142) + log(60.9)] x 10-7
= antilog [0.6021 + 0.4972 + 1.7846] x 10-7
= antilog [2.8839] x 10-7
= 765.4 x 10-7
∴ μ = 7.654 x 10-5 Wb/A-m
The magnectic permeability of the iron rod is 7.654 x 10-5 Wb/A-m.
Concept: Magnetic Properties of Materials
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