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)

7 Views | Posted 4 months ago

Asked by Shiksha User

1 Answer

Answered by

Payal Gupta | Contributor-Level 10

4 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>

0 Upvotes 0 Downvotes

1 Answer

Similar Questions for you

Learn more about...

Most viewed information

Share Your College Life Experience

Didn't find the answer you were looking for?

Need guidance on career and education? Ask our experts

Your Question