JEE MAIN - Physics (2020 - 4th September Evening Slot - No. 24)
A paramagnetic sample shows a net magnetisation of 6 A/m when it is placed in an external
magnetic field of 0.4 T at a temperature of 4 K. When the sample is placed in an external magnetic
field of 0.3 T at a temperature of 24 K, then the magnetisation will be:
4 A/m
1 A/m
0.75 A/m
2.25 A/m
Explanation
According to curies law
$$\chi $$ $$ = {{C{B_{ext}}} \over T}$$
$$ \Rightarrow $$ $$6 = {{C \times 0.4} \over 4}$$
$$ \Rightarrow C = 60$$
$$ \therefore $$ Case - II : $$\chi $$ $$ = {{60 \times 0.3} \over {24}}$$ $$ = {{60 \times 3} \over {240}} = {3 \over 4} = 0.75\,A/m$$
$$\chi $$ $$ = {{C{B_{ext}}} \over T}$$
$$ \Rightarrow $$ $$6 = {{C \times 0.4} \over 4}$$
$$ \Rightarrow C = 60$$
$$ \therefore $$ Case - II : $$\chi $$ $$ = {{60 \times 0.3} \over {24}}$$ $$ = {{60 \times 3} \over {240}} = {3 \over 4} = 0.75\,A/m$$
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