An iron rod of the area of cross-section 0.1m^{2} is subjected to a magnetizing 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]

An iron rod of the area of cross-section 0.1 m^{2} is subjected to a magnetizing field of 1000 A/m. Calculate the magnetic permeability of the iron rod.

(χ for iron = 599 , µ_{0} = 4π × 10^{-7} SI unit)

#### Solution 1

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.**

#### Solution 2

**Given:**

H = 1000 A/m, χ = 599,

μ_{0} = 4π × 10^{-7} S.I. unit

**To find:** Permeability (μ)

**Formula: **μ = μ_{0}(1 + χ)

**Calculation: **

From formula,

μ = 4π × 10^{-7 }(1 + 599)

= 4 × 3.142 × 10^{-7} × 600

∴ μ = 7.54 × 10^{-4} Hm^{-1 }

**The magnetic permeability of the iron rod is 7.54 × 10 ^{-4} Hm^{-1}. **