Ncert Exercises & Solutions

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)

Number of atomic dipoles,

n = 2.0 \times 10^{24}

Dipole moment of each atomic dipole,

M = 1.5 \times 10^{- 23} J T^{- 1}

When the magnetic field,

B_{1} = 0.64 T

The sample is cooled to 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^{- 1}

Magnetic saturation is achieved at 15%.
15
Hence, effective dipole moment,

M_{1} = \frac{15}{100} \times 0 = 4.5 j T^{- 1}

When the magnetic field,

B_{2} = 0.98 T

Temperature,

T_{2} = 2.8 ° K

ts total dipole moment = M2
According to Curie's law,
we have the ratio of two magnetic dipoles as:

\begin{aligned} \frac{M_{2}}{M_{1}} & = \frac{B_{2}}{B_{1}} \times \frac{T_{1}}{T_{2}} \\ ∴ M_{2} & = \frac{B_{2} T_{1} M_{1}}{B_{1} T_{2}} \\ = \frac{0.98 \times 4.2 \times 4.5}{2.8 \times 0.64} = 10.336 {JT}^{- 1} \end{aligned}

Therefore,

10.336 j T^{- 1}

Is the total dipole moment of the sample for a magnetic field of 0.98 T and a temperature of 2.8K.

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