Each atom of an iron bar `(5cmxx1cmxx1cm)` has a magnetic moment `1.8xx10^(-23) Am^(2)` Knowing that the density of iron is `7.78xx10^(3) kg^(-3)m` atomic weight is `56` and Avogadro's number is `6.02xx10^(23)` the magnetic moment of bar in the state of magnetic saturation will be

Options

A. `4.75 Am^(2)`
B. `5.75 Am^(2)`
C. `75.4 Am^(2)`
D. `75.4 Am^(2)`

Answer

C

Explanation

The number of atoms per unit volume in a specimen
`n=(rhoN_(A))/A`
for iron, `rho=7.8xx10^(-3) kgm^(-3)`,
`N_(A)=6.02xx10^(26)//kg mol, A=56`
`impliesn=(7.8xx10^(3)xx6.02xx10^(26))/56=8.38xx10^(28)m^(-3)`
Total number of atoms in the bar is
`N_(0)=nV=8.38xx10^(28)xx(5xx10^(-2)xx1xx10^(-2)xx1xx10^(-2))`
`N_(0)=4.19xx10^(23)`
The saturated magnetic moment of bar
`=4.19xx10^(23)xx1.8xx10^(-23)=7.54 Am^(2)`