The Curie temperatures of cobalt and iron are | vedclass

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(B) According to the Curie-Weiss law, the magnetic susceptibility $\chi$ of a ferromagnetic material above its Curie temperature $T_C$ is given by $\c

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$3$

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(B) According to the Curie-Weiss law, the magnetic susceptibility $\chi$ of a ferromagnetic material above its Curie temperature $T_C$ is given by $\chi = \frac{C}{T - T_C}$, where $C$ is the Curie constant and $T$ is the absolute temperature.Since the Curie constant $C$ is approximately the same for these materials, we have $\chi \propto \frac{1}{T - T_C}$.Therefore, the ratio of the magnetic susceptibility of cobalt to that of iron is given by:$\frac{\chi_{	ext{cobalt}}}{\chi_{	ext{iron}}} = \frac{T - (T_C)_{	ext{iron}}}{T - (T_C)_{	ext{cobalt}}}$Given $T = 1600 \,K$, $(T_C)_{	ext{cobalt}} = 1400 \,K$, and $(T_C)_{	ext{iron}} = 1000 \,K$.Substituting these values:$\frac{\chi_{	ext{cobalt}}}{\chi_{	ext{iron}}} = \frac{1600 - 1000}{1600 - 1400} = \frac{600}{200} = 3$.

The Curie temperatures of cobalt and iron are $1400 \,K$ and $1000 \,K$ respectively. At $T=1600 \,K$, the ratio of the magnetic susceptibility of cobalt to that of iron is

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