NEET-XII-Physics
37: Magnetic Properties of Matter
- #6The 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 µr and the susceptibility χ of the material.Ans : 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
`` \mu \mathit{=}\frac{B}{H}``
`` \Rightarrow \mu =\frac{1.6}{1000}``
`` \Rightarrow \mu =1.6\times {10}^{-3}``
`` \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}``
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