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For a magnetising field of intensity 2xx...

For a magnetising field of intensity `2xx10^3A//m`, aluminium at `280K` acquires intensity of magnetisation of `4*8xx10^-2Am^-1`. Find the susceptibility of aluminium at `280K`. If the temperature of the metal is raised to `320K`, what will be its susceptibility and intensity of magnetisation?

Text Solution

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Here, `H=2xx10^-3A//m`, `T=280K`
`I=4*8xx10^-2Am^-1`, `chi_m=?`
`chi_m=I/H=(4*8xx10^-2)/(2xx10^3)=2*4xx10^-5`
Now, `T^'=320K`, `chi_(m)^'=?` `I^'=?`
According to Curie's Law.
`chi_m^'=chi_mxxT/T^'=2*4xx10^-5xx(280)/(320)=2*1xx10^-5`
The new intensity of magnetisation will be from
`chi_m^'=I^'/H`
`I^'=Hchi_m^'=2xx10^3xx2*1xx10^-5`
`I^'=4*2xx10^-2Am^-1`
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