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.

Answer 2

The problem asks for the ratio of the magnetic susceptibility of Cobalt (\(\chi_{Co}\)) to that of Iron (\(\chi_{Fe}\)) at a temperature \(T = 1600 \, K\).

We are given the Curie temperatures for Cobalt and Iron:

• Curie temperature of Cobalt, \(T_{c,Co} = 1400 \, K\)
• Curie temperature of Iron, \(T_{c,Fe} = 1000 \, K\)

The given temperature \(T = 1600 \, K\) is above the Curie temperature for both Cobalt (\(T > T_{c,Co}\)) and Iron (\(T > T_{c,Fe}\)). Above the Curie temperature, ferromagnetic materials behave like paramagnetic materials, and their magnetic susceptibility (\(\chi\)) follows the Curie-Weiss law: \[ \chi = \frac{C}{T - T_c} \] where \(C\) is the Curie constant specific to the material, \(T\) is the absolute temperature, and \(T_c\) is the Curie temperature.

For Cobalt at \(T = 1600 \, K\): \[ \chi_{Co} = \frac{C_{Co}}{T - T_{c,Co}} = \frac{C_{Co}}{1600 - 1400} = \frac{C_{Co}}{200} \]

For Iron at \(T = 1600 \, K\): \[ \chi_{Fe} = \frac{C_{Fe}}{T - T_{c,Fe}} = \frac{C_{Fe}}{1600 - 1000} = \frac{C_{Fe}}{600} \]

The ratio of the magnetic susceptibilities is: \[ \frac{\chi_{Co}}{\chi_{Fe}} = \frac{\left(\frac{C_{Co}}{200}\right)}{\left(\frac{C_{Fe}}{600}\right)} = \frac{C_{Co}}{C_{Fe}} \times \frac{600}{200} = \frac{C_{Co}}{C_{Fe}} \times 3 \]

Assuming the Curie constants for Cobalt and Iron are approximately equal (\(C_{Co} \approx C_{Fe}\)), which is often a simplifying assumption in such problems when the constants are not provided and simple numerical options are given, the ratio becomes: \[ \frac{\chi_{Co}}{\chi_{Fe}} \approx 1 \times 3 = 3 \]

Thus, the ratio of magnetic susceptibility of Cobalt to that of Iron at \(T = 1600 \, K\) is approximately 3.

The calculated ratio matches option (A).