Question

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

• A. 3
• B. \(\frac{7}{5}\)
• C. \(\frac{5}{7}\)
• D. \(\frac{1}{3}\)

Answer 1

The Correct Option is A

Given: 
Curie temperature of Cobalt, \( T_{C(Co)} = 1400\,K \)
Curie temperature of Iron, \( T_{C(Fe)} = 1000\,K \)
Temperature at which susceptibility is measured, \( T = 1600\,K \)

Step-by-Step Explanation:

Step 1: Use Curie-Weiss law for magnetic susceptibility (\( \chi \)) above Curie temperature (\( T > T_C \)):

According to Curie-Weiss law: \[ \chi = \frac{C}{T - T_C} \]

where \( C \) is the Curie constant.

Step 2: Calculate magnetic susceptibility for Cobalt at \( T = 1600\,K \):

\[ \chi_{Co} = \frac{C_{Co}}{T - T_{C(Co)}} = \frac{C_{Co}}{1600 - 1400} = \frac{C_{Co}}{200} \]

Step 3: Calculate magnetic susceptibility for Iron at \( T = 1600\,K \):

\[ \chi_{Fe} = \frac{C_{Fe}}{T - T_{C(Fe)}} = \frac{C_{Fe}}{1600 - 1000} = \frac{C_{Fe}}{600} \]

Important Note: The Curie constant \( C \) depends on material-specific factors. However, since Cobalt and Iron are both ferromagnetic materials with similar magnetic behaviors, we assume their Curie constants (\( C_{Co} \approx C_{Fe} \)) are nearly equal for comparison.

Step 4: Now calculate the ratio \(\frac{\chi_{Co}}{\chi_{Fe}}\):

Assuming \( C_{Co} \approx C_{Fe} \):

\[ \frac{\chi_{Co}}{\chi_{Fe}} = \frac{\frac{C_{Co}}{200}}{\frac{C_{Fe}}{600}} = \frac{600}{200} = 3 \]

Thus, the ratio of magnetic susceptibility of Cobalt to Iron is 3.