5.23 A sample of paramagnetic salt contains 2.0 × 10²⁴ atomic dipoles each of dipole moment 1.5 × 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)

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Payal Gupta | Contributor-Level 10

5 months ago

5.23 Number of atomic dipoles, n = 2 $\times 10^{24}$

Dipole moment of each atomic dipole, M = 1.5 $\times 10^{- 23}$ J/T

Magnetic field, $B_{1}$ = 0.64 T

The sample is cooled to the temperature, $T_{1}$ = 4.2 K

Total dipole moment of the atomic dipole, $M_{t o t}$ = n $\times$ M

= 2 $\times 10^{24}$ $\times 1.5 \times 10^{- 23} = 30 J / T$

Magnetic saturation is achieved at 15 %

Hence, effective dipole moment, = 15 % $\times 30$ = 4.5 J/T

When the magnetic field, $B_{2}$ = 0.98 T, Temperature, $T_{2}$ = 2.8 K and total dipole moment = $M_{2}$

According to Curie’s law, the ratio of the two dipole moment

$\frac{M_{2}}{M_{1}}$ = $\frac{B_{2}}{B_{1}} \times \frac{T_{1}}{T_{2}}$

$M_{2}$ = $\frac{M_{1} \times B_{2} \times T_{1}}{B_{1} \times T_{2}}$ = $\frac{4.5 \times 0.98 \times 4.2}{0.64 \times 2.8}$ = 10.336 J/T

Therefore, 10.336 J/T is the total dipole moment for a magnetic field of 0.98 T at a temperature of 2.8 K.

5.23 Number of atomic dipoles, n = 2 <math> <mo>×</mo> <mi> </mi> <msup> <mrow> <mrow> <mn>10</mn> </mrow> </mrow> <mrow> <mrow> <mn>24</mn> </mrow> </mrow> </msup> </math> Dipole moment of each atomic dipole, M = 1.5 <math> <mo>×</mo> <msup> <mrow> <mrow> <mn>10</mn> </mrow> </mrow> <mrow> <mrow> <mo>-</mo> <mn>23</mn> </mrow> </mrow> </msup> </math> J/TMagnetic field, <math> <msub> <mrow> <mrow> <mi>B</mi> </mrow> </mrow> <mrow> <mrow> <mn>1</mn> </mrow> </mrow> </msub> </math> = 0.64 TThe sample is cooled to the temperature, <math> <msub> <mrow> <mrow> <mi>T</mi> </mrow> </mrow> <mrow> <mrow> <mn>1</mn> </mrow> </mrow> </msub> </math> = 4.2 KTotal dipole moment of the atomic dipole, <math> <msub> <mrow> <mrow> <mi>M</mi> </mrow> </mrow> <mrow> <mrow> <mi>t</mi> <mi>o</mi> <mi>t</mi> </mrow> </mrow> </msub> </math> = n <math> <mo>×</mo> </math> M= 2 <math> <mo>×</mo> <mi> </mi> <msup> <mrow> <mrow> <mn>10</mn> </mrow> </mrow> <mrow> <mrow> <mn>24</mn> </mrow> </mrow> </msup> </math> <math> <mo>×</mo> <mi> </mi> <mn>1.5</mn> <mi mathvariant="normal"> </mi> <mo>×</mo> <msup> <mrow> <mrow> <mn>10</mn> </mrow> </mrow> <mrow> <mrow> <mo>-</mo> <mn>23</mn> </mrow> </mrow> </msup> <mi mathvariant="normal"> </mi> <mo>=</mo> <mi mathvariant="normal"> </mi> <mi mathvariant="normal"> </mi> <mn>30</mn> <mi mathvariant="normal"> </mi> <mi mathvariant="normal"> </mi> <mi mathvariant="normal">J</mi> <mo>/</mo> <mi mathvariant="normal">T</mi> </math> Magnetic saturation is achieved at 15 %Hence, effective dipole moment, = 15 % <math> <mo>×</mo> <mn>30</mn> </math> = 4.5 J/T<div>When the magnetic field, <math> <msub> <mrow> <mrow> <mi>B</mi> </mrow> </mrow> <mrow> <mrow> <mn>2</mn> </mrow> </mrow> </msub> </math> = 0.98 T, Temperature, <math> <msub> <mrow> <mrow> <mi>T</mi> </mrow> </mrow> <mrow> <mrow> <mn>2</mn> </mrow> </mrow> </msub> </math> = 2.8 K and total dipole moment = <math> <msub> <mrow> <mrow> <mi>M</mi> </mrow> </mrow> <mrow> <mrow> <mn>2</mn> </mrow> </mrow> </msub> </math> According to Curie’s law, the ratio of the two dipole moment <math> <mfrac> <mrow> <mrow> <msub> <mrow> <mrow> <mi>M</mi> </mrow> </mrow> <mrow> <mrow> <mn>2</mn> </mrow> </mrow> </msub> </mrow> </mrow> <mrow> <mrow> <msub> <mrow> <mrow> <mi>M</mi> </mrow> </mrow> <mrow> <mrow> <mn>1</mn> </mrow> </mrow> </msub> </mrow> </mrow> </mfrac> </math> = <math> <mfrac> <mrow> <mrow> <msub> <mrow> <mrow> <mi>B</mi> </mrow> </mrow> <mrow> <mrow> <mn>2</mn> </mrow> </mrow> </msub> </mrow> </mrow> <mrow> <mrow> <msub> <mrow> <mrow> <mi>B</mi> </mrow> </mrow> <mrow> <mrow> <mn>1</mn> </mrow> </mrow> </msub> </mrow> </mrow> </mfrac> <mo>×</mo> <mfrac> <mrow> <mrow> <msub> <mrow> <mrow> <mi>T</mi> </mrow> </mrow> <mrow> <mrow> <mn>1</mn> </mrow> </mrow> </msub> </mrow> </mrow> <mrow> <mrow> <msub> <mrow> <mrow> <mi>T</mi> </mrow> </mrow> <mrow> <mrow> <mn>2</mn> </mrow> </mrow> </msub> </mrow> </mrow> </mfrac> </math> <math> <msub> <mrow> <mrow> <mi>M</mi> </mrow> </mrow> <mrow> <mrow> <mn>2</mn> </mrow> </mrow> </msub> </math> = <math> <mfrac> <mrow> <mrow> <msub> <mrow> <mrow> <mi>M</mi> </mrow> </mrow> <mrow> <mrow> <mn>1</mn> </mrow> </mrow> </msub> <mo>×</mo> <msub> <mrow> <mrow> <mi>B</mi> </mrow> </mrow> <mrow> <mrow> <mn>2</mn> </mrow> </mrow> </msub> <mo>×</mo> <msub> <mrow> <mrow> <mi>T</mi> </mrow> </mrow> <mrow> <mrow> <mn>1</mn> </mrow> </mrow> </msub> </mrow> </mrow> <mrow> <mrow> <msub> <mrow> <mrow> <mi>B</mi> </mrow> </mrow> <mrow> <mrow> <mn>1</mn> </mrow> </mrow> </msub> <mo>×</mo> <msub> <mrow> <mrow> <mi>T</mi> </mrow> </mrow> <mrow> <mrow> <mn>2</mn> </mrow> </mrow> </msub> </mrow> </mrow> </mfrac> </math> = <math> <mfrac> <mrow> <mrow> <mn>4.5</mn> <mi mathvariant="normal"> </mi> <mo>×</mo> <mn>0.98</mn> <mo>×</mo> <mn>4.2</mn> </mrow> </mrow> <mrow> <mrow> <mn>0.64</mn> <mo>×</mo> <mn>2.8</mn> </mrow> </mrow> </mfrac> </math> = 10.336 J/TTherefore, 10.336 J/T is the total dipole moment for a magnetic field of 0.98 T at a temperature of 2.8 K.</div>

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