KCET · Physics · Magnetic Properties of Matter
A paramagnetic sample shows a net magnetisation of \(8 \mathrm{Am}^{-1}\) when placed in an external magnetic field of \(0.6 \mathrm{~T}\) at a temperature of \(4 \mathrm{~K}\). When the same sample is placed in an external magnetic field of \(0.2 \mathrm{~T}\) at a temperature of \(16 \mathrm{~K}\), the magnetisation will be
- A \(\frac{32}{3} \mathrm{Am}^{-1}\)
- B \(\frac{2}{3} \mathrm{Am}^{-1}\)
- C \(6 \mathrm{Am}^{-1}\)
- D \(2.4 \mathrm{Am}^{-1}\)
Answer & Solution
Correct Answer
(B) \(\frac{2}{3} \mathrm{Am}^{-1}\)
Step-by-step Solution
Detailed explanation
Given, \(M_{1}=8 \mathrm{Am}^{-1}, B_{1}=0.6 \mathrm{~T}, T_{1}=4 \mathrm{~K}\), \(B_{2}=0.2 \mathrm{~T}\) and \(T_{2}=16 \mathrm{~K}\)
According to Curie's law, for paramagnetic materials,
\(M=\frac{C B}{T}\)
\(\Rightarrow M_{1}=\frac{C B_{1}}{T_{1}}\)
and \(M_{2}=\frac{C B_{2}}{T_{2}}\)
\(\therefore \frac{M_{2}}{M_{1}} \frac{\frac{C B_{2}}{T_{2}}}{\frac{C B_{1}}{T_{1}}}=\frac{B_{2}}{B_{1}} \times \frac{T_{1}}{T_{2}}\)
\(=\frac{0.2}{0.6} \times \frac{4 \mathrm{~K}}{16 \mathrm{~K}}=\frac{1}{3} \times \frac{1}{4}\)
\(\Rightarrow \frac{M_{2}}{M_{1}}=\frac{1}{12} \Rightarrow M_{2}=\frac{M_{1}}{12}=\frac{8}{12}\)
\(\Rightarrow M_{2}=\frac{2}{3} \mathrm{Am}^{-1}\)
According to Curie's law, for paramagnetic materials,
\(M=\frac{C B}{T}\)
\(\Rightarrow M_{1}=\frac{C B_{1}}{T_{1}}\)
and \(M_{2}=\frac{C B_{2}}{T_{2}}\)
\(\therefore \frac{M_{2}}{M_{1}} \frac{\frac{C B_{2}}{T_{2}}}{\frac{C B_{1}}{T_{1}}}=\frac{B_{2}}{B_{1}} \times \frac{T_{1}}{T_{2}}\)
\(=\frac{0.2}{0.6} \times \frac{4 \mathrm{~K}}{16 \mathrm{~K}}=\frac{1}{3} \times \frac{1}{4}\)
\(\Rightarrow \frac{M_{2}}{M_{1}}=\frac{1}{12} \Rightarrow M_{2}=\frac{M_{1}}{12}=\frac{8}{12}\)
\(\Rightarrow M_{2}=\frac{2}{3} \mathrm{Am}^{-1}\)
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