Question

The magnetic field B and the magnetic intensity H in a material are found to be 1.6 T and 1000 A m⁻¹, respectively. Calculate the relative permeability µ_r and the susceptibility χ of the material.

Answer 1

Here,
Magnetic field strength, B = 1.6 T
Magnetising intensity in a material, H = 1000 A/m
The relation between magnetic field and magnetising field is given by
$$\begin{gathered} \mu \mathit{=}\frac{B}{H} \\ \Rightarrow \mu =\frac{1.6}{1000} \\ \Rightarrow \mu =1.6\times {10}^{-3} \end{gathered}$$
$$\begin{gathered} \text{R}\mathrm{elative}\mathrm{perm}\text{ea}\mathrm{bility}\mathrm{is}\mathrm{defined}\mathrm{as}\mathrm{the}\mathrm{ratio}\mathrm{of}\mathrm{permeability}\mathrm{in}\text{a}\mathrm{medium}\mathrm{to}\mathrm{that}\mathrm{in}\mathrm{vacuum}. \\ \text{So,}{\mu }_r=\frac{\mu }{{\mu }_0}=\frac{1.6\times {10}^{-3}}{4\mathrm{\pi }\times {10}^{-7}} \\ \Rightarrow {\mu }_r=0.127\times {10}^4 \\ \Rightarrow {\mu }_r=1.3\times {10}^3 \\ \mathrm{Relative}\mathrm{permeability}\left({\mu }_r\right)\mathrm{is}\mathrm{given}\mathrm{by}, \\ {\mu }_r=(1+\chi ) \\ \Rightarrow \chi =1.3\times {10}^3-1 \\ \Rightarrow \chi =1300-1 \\ \Rightarrow \chi =1299=1.299\times {10}^3 \\ \Rightarrow \chi \approx 1.3\times {10}^3 \end{gathered}$$