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)

Answers (1)
H Harsh Kankaria

Given,

Magnetic field, B_{1} = 0.64 T

Temperature, \theta_{1} = 4.2K

And, saturation = 15%

Hence, Effective dipole moment, M_{1} = 15% of Total dipole moment

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

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

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

Let M_{2} be the new dipole moment.

We know that according to Curie’s Law, 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} = 10.336JT^{-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 JT^{-1}

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