A magnetising field of `1500A m^(-1)` produces flux of `2.4xx 10^(-5)` weber in a iron bar of the cross-sectional area of `0.5cm^(2)`. The permeability of the iron bar is

Answer

A, B, D

Explanation

`B=phi/A=(2.4xx10^(-5) Wb)/(0.2xx10^(-4)m^(2))=1.2 Wb//m^(2)=1.2NA^(-1)m^(-1)`
The magnetising field (or megnetic intensity) `H` is `1600 Am^(-1)`.Therefore, the magnetic permeability is given by-
`mu=B/H=(1.2NA^(-1)m^(-1))/(1600 Am^(-1))=7.5xx10^(-4)N//A^(2)`.
Now,from the relation `mu=mu_(0)(1+chi_(m))`,the susceptibility is given by
`chi_(m)=mu/mu_(0)-1`
We known that `mu_(0)=4pixx10^(-7)N//A^(2)`.
`:.chi_(m)=(7.5xx10^(-4))/(4xx3.14xx10^(-7))-1=596`.