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A paramagnetic sample shows a net magnetization of 0.8 A/m, when placed in an external magnetic field of strength 0.8T at a temperature 5K. When the same sample is placed in an external magnetic field of 0.4T at a temperature of 20K, the magnetization is

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0.8 Am^–1

0.4 Am^–2

0.2 Am^-1

0.1 Am^–1

answer is D.

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Detailed Solution

$χ = \frac{I}{H} = \frac{C}{T} {where B}_{ext} = μ_{o} H \Rightarrow \frac{IT}{B_{ext}} = \frac{C}{μ_{o}} = constant$

Thus,

$\frac{I_{1}}{I_{2}} = \frac{B_{1} \times T_{2}}{B_{2} \times T_{1}} \frac{0.8}{I_{2}} = \frac{0.8 \times 20}{0.4 \times 5} \Rightarrow I_{2} = \frac{0.4 \times 5}{20} = 0.1 {Am}^{- 1}$

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