Question

A magnetic field of 1800A/m produces a magnetic flux 3X10-5 wb/m2 in an iron bar of cross sectional area 0.2 cm2, calculate the permeability and relative permittivity of the material.

Answer 1

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Answer 2

The permeability of the material is 8.33 × 10⁻⁴

The relative permittivity of the material is 663.1

Explanation:

Given:

Magnetic Field H = 1800 A/m

Magnetic Flux Φ = 3 × 10⁻⁵ Wb/m²

Cross sectional area of the iron bar A = 0.2 cm²

To find out:

Permeability and relative permeability of the material

Solution:

Magnetic Flux Density

$B=\frac{\phi}{A}$

$\implies B=\frac{3\times 10^{-5}}{0.2\times 10^{-4}}$

$\implies B=1.5$

Therefore,

Permeability

$\mu=\frac{B}{H}$

$\implies \mu=\frac{1.5}{1800}$

$\implies \mu=8.33\times 10^{-4}$

Relative permeability

$\mu_r=\frac{\mu}{\mu_0}$

$\implies \mu_r=\frac{8.33\times 10^{-4}}{4\pi\times 10^{-7}}$

$\implies \mu_r=0.6631\times 10^3=663$

Hope this answer is helpful.

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