• # 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,             $I'=2.1\times {{10}^{-5}}\times 2\times {{10}^{3}}=4.2\times {{10}^{-2}}A{{m}^{-1}}$

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