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Question

Physics Question on Moving charges and magnetism

The relation connecting magnetic susceptibility χm\chi_{m} and relative permeability μr\mu _{r} is

A

χm=μr+1\chi _{m}=\mu_{r}+1

B

χm=μr1\chi _{m}=\mu _{r}-1

C

χm=1μr\chi _{m}=\frac{1}{\mu _{r}}

D

χm=3(1+μr)\chi _{m}=3(1+\mu _{r})

Answer

χm=μr1\chi _{m}=\mu _{r}-1

Explanation

Solution

Total magnetic flux density BB in a material is the sum of magnetic flux density in vacuum B0B_{0} produced by magnetising force and magnetic flux density due to magnetisation of material Bm B_{m} ie, B=B0+BmB=B_{0}+B_{m} \Rightarrow B=μ0H+μ0I=μ0(H+I)B=\mu _{0}H+\mu_{0}I=\mu _{0}(H+I) =μ0H(1+χm)=\mu _{0}H(1+\chi _{m}) Also μr=(1+χm)\mu _{r}=(1+\chi _{m})