A sample of paramagnetic salt contains `2xx10^(24)` atomic dipoles, each of moment `1.5xx10^(-23)JT^-1`. The sample is placed under a homogeneous magnetic field of `0.64T` and cooled to a temperature of `4.2K`. The degree of magnetic saturation archieved is equal to `15%`. What is the total dipole moment of the sample for a mangetic field of `0.98T` and a temperature of `2.8K`. (Assume Curie's law).

Answer

`7.9`

Explanation

According to Curie.s law , `chi_(m) = C/T `
As `chi_(m) =M/H , M =m/V and H = B/mu`
`(m//V)/(B//mu)=C/T`
or `m =(CV)/mu (B/T)`
For a given sample , `CV//mu` = constant
Thus , `mprop (B//T) or (m_(1))/(m_(2)) =(B_(1)//T_(1))/(B_(2)//T_(2))`
We are given that , `B_(1) = 0.84 T , B_(2) = 0.98 T ` ,
`T_(1)=4.2 K and T_(2) = 2.8` K
Thus , `(m_(1))/(m_(2)) = (0.84//4.2 )/(0.98//2.8)=4/7`
or `m_(2) = (7//4)m_(1)`
As per the given conditions ,
Initial total magnetic moment of the sample ,i.e.,
`m_(1)` = 15 % of `(2.0 xx 10^(24)) (1.5 xx 10^(-23)J//T)=4.5 ` J/T
Thus , `m_(2)`= (7/4)(4.5 J/T) =7.9 J/T
Thus , `m_(2)` = (7/4) (4.5J/T) = 7.9 J/T