An iron bar of length `10 cm` and diameter `2 cm` is placed in a magnetic field of intensity `1000Am^(-1)` with its length parallel to the direction of the field. Determine the magnetic moment produced in the bar if permeability of its material is `6.3xx10^(-4) TmA(-1)`.

Explanation

we know that, `mu=mu_(0)(1+chi)`
`rArrchi=mu/mu_(0)-1=(6.3xx10^(-4))/(4pixx10^(-7))-1=500.6`
Intensity of magnetisation,
`I=chiH=500.6xx1000=5xx10^(5)Am^(-1)`
`therefore` magnetic moment, `M=IxxV=Ixxpir^(2)l`
= `5xx10^(5)xx3.14xx(10^(-2))^(2)xx(10xx10^(-2))=17.70A-m^(2)`