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 `A m^(2)`
B. 5,74 `A m^(2)`
C. 7.54 `A m^(2)`
D. 75.4 `A m^(2)`

Answer

C

Explanation

No. of atom per unit volume
`n= (rho N_(A))/(A)= (7.8 xx 10^(3) xx 6.02 xx 10^(23))/(56 xx 10^(-3))= 8.38 xx 10^(28)//m^(3)`
No. of atom
`N_(0)= nV = 8.38 xx 10^(28) xx (5 xx 10^(-2) xx 1 xx 10^(-2) xx 1 xx 10^(-2))= 4.19 xx 10^(23)`
Magnetic moment at magnetic saturation state `= N_(0) M_(0)= 4.19 xx 10^(23) xx 1.8 xx 10^(-23) = 7.54 A m^(2)`