# 5.23. A sample of paramagnetic salt contains $2.0 \times 10 ^{24}$atomic dipoles each of dipole moment $1.5 \times 10 ^{-23} JT^{-1}$ 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)

Given,

Magnetic field, $\dpi{100} B_{1}$ = 0.64 T

Temperature, $\dpi{100} \theta_{1}$ = 4.2K

And, saturation = 15%

Hence, Effective dipole moment, $\dpi{80} M_{1}$ = 15% of Total dipole moment

$\dpi{80} M_{1}$ = 0.15 x (no. of atomic dipole × individual dipole moment)

$\dpi{100} M_{1}$ = $\dpi{100} 0.15 \times 2 \times 10^{24} \times 1.5 \times 10^{-23}$= 4.5 $\dpi{80} JT^{-1}$

Now,  Magnetic field, $\dpi{100} B_{2}$ = 0.98 T and Temperature, $\dpi{100} \theta_{2}$ = 2.8 K

Let $\dpi{80} M_{2}$ be the new dipole moment.

We know that according to Curie’s Law, $\dpi{80} M\propto \frac{B}{\theta}$

∴ The ratio of magnetic dipole moments

$\\\frac{M_{2}}{M_{1}}= \frac{B_{2}\times \theta_{1}}{B_{1}\times \theta_{2}} \\ \implies M_{2}= \frac{B_{2}\times \theta_{1}}{B_{1}\times \theta_{1}}\times M_{1} \\\implies M_{2}= \frac{0.98T\times4.2K}{0.64T\times2.8K}\times 4.5 JT^{-1}$ $= 10.336$$JT^{-1}$

Therefore, the total dipole moment of the sample for a magnetic field of 0.98 T and a temperature of 2.8 K = 10.336 $\dpi{80} JT^{-1}$

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