question_answer

For a magnetising field of intensity \[2\times {{10}^{3}}A{{m}^{-1}},\] aluminium at 280 K acquires intensity of magnetisation of \[4.8\times {{10}^{-2}}{{m}^{-1}}A.\] Find the susceptibility of aluminium at 280 K. If the temperature of the metal is raised to 320 K, then what will be its susceptibility and intensity of magnetisation?

Answer:

            Here, \[H=2\times {{10}^{3}}A{{m}^{-1}},\] \[T=280\,K\]             \[I=4.8\times {{10}^{-2}}A{{m}^{-1}}\] \[\therefore \]Susceptibility\[{{\chi }_{m}}=\frac{I}{H}=\frac{4.8\times {{10}^{-2}}}{2\times {{10}^{3}}}=2.4\times {{10}^{-5}}\]             \[\chi {{'}_{m}}=?,\] \[T'=320K\] According to Curie?s law, \[\frac{\chi {{'}_{m}}}{{{\chi }_{m}}}=\frac{T}{T'}\Rightarrow \chi {{'}_{m}}=\frac{T}{T'}\times {{\chi }_{m}}\] or         \[\chi {{'}_{m}}=\frac{280}{320}\times 2.4\times {{10}^{-5}}=2.1\times {{10}^{-5}}\] As,        \[\chi {{'}_{m}}=\frac{I'}{H}\Rightarrow I'=\chi 'm\times H\] or Intensity of magnetization,             \[I'=2.1\times {{10}^{-5}}\times 2\times {{10}^{3}}=4.2\times {{10}^{-2}}A{{m}^{-1}}\]