Question

A sample of paramagnetic salt contains $2.0\times {10}^{24}$ atomic dipoles each of dipole moment $1.5\times {10}^{-23}J{T}^{-1}$. The sample is placed under a homogeneous magnetic field of $0.64$ T, and cooled to a temperature of $4.2$ K. The degree of magnetic saturation achieved is equal to $15$%. What is the total dipole moment of the sample for a magnetic field of $0.98$ T and a temperature of $2.8$ K? (Assume Curie's law)

Answer 1

Given that,

Number of dipole = 2×10²⁴

Dipole moment of each dipole = 1.5×10^-23 J/T

Degree of magnetic saturation = 15%

Homogeneous magnetic field B₁ = 0.64 T

Initial temperature T₁ = 4.2 K

Final magnetic field B₂ = 0.98 T

Final temperature T₂ = 2.8 K

Dipole moment of sample M = Number of dipoles × dipole moment of each dipole

$M=2\times {10}^{24}\times 1.5\times {10}^{-23}$

$M=30$

At saturation M₁ = 15% M

$M_1=\frac{15}{100}\times 30$

$M_1=4.5J/T$

According to the curies law

$\frac{C}{T}=\frac{T}{H}$

$T\propto MandH\propto B$

Now,

$\frac{C}{T}=\frac{M}{B}$

$M\propto \frac{B.C}{T}$

Now,

$\frac{M_1}{M_2}=\frac{B_1}{T_1}\times \frac{T_2}{B_2}$

$M_2=\frac{B_2{T}_1}{B_1{T}_2}\times M_1$

$M_2=\frac{0.98\times 4.2}{0.64\times 2.8}\times 4.5$

$M_2=10.335J/T$

$M_2=10.34J/T$

Hence, the dipole moment of the sample is$10.34$ J/T.