AP EAMCET · PHYSICS · Magnetic Properties of Matter
A sample of paramagnetic salt contains \(2 \times 10^{24}\) atomic dipoles each of dipole moment \(1.5 \times 10^{-23} \mathrm{JT}^{-1}\). The sample is placed under homogeneous magnetic field of 0.6 T and cooled to a temperature 4.2 K. The degree of magnetic saturation achived is \(20 \%\). Then total dipole moment of the sample for a magnetic field of 0.9 T and a temperature of 2.8 K is
- A \(4.5 \mathrm{~J} \mathrm{~T}^{-1}\)
- B \(13.5 \mathrm{~J} \mathrm{~T}^{-1}\)
- C \(0.64 \mathrm{~J} \mathrm{~T}^{-1}\)
- D \(7 \mathrm{~J} \mathrm{~T}^{-1}\)
Answer & Solution
Correct Answer
(B) \(13.5 \mathrm{~J} \mathrm{~T}^{-1}\)
Step-by-step Solution
Detailed explanation
\( M_{sat} = N \mu = (2 \times 10^{24})(1.5 \times 10^{-23} \mathrm{JT}^{-1}) = 30 \mathrm{JT}^{-1} \) \( M_1 = 0.20 M_{sat} = 0.20 \times 30 \mathrm{JT}^{-1} = 6 \mathrm{JT}^{-1} \) \( \frac{M_1 T_1}{B_1} = \frac{M_2 T_2}{B_2} \)…
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