#### Question

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, μ_{0} = 4π x 10^{-7} S.I. unit

To find:- Permeability (μ)

Formula:- μ = μ_{0} (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.**

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Solution for question: Calculate the Magnetic Permeability of the Iron Rod concept: Magnetic Properties of Materials. For the courses HSC Science (Computer Science), HSC Science (Electronics), HSC Science (General)