Question

A paramagnetic sample shows a net magnetization of $8\, Am^{-1}$ when placed in an external magnetic field of $0.6\,T$ at a temperature of $4\,K$. When the same sample is placed in an external magnetic field of $0.2 \,T$ at a temperature of $16\, K$. the magnetization will be

• A. $\frac{32}{3} Am^{-1}$
• B. $\frac{2}{3} Am^{-1}$
• C. $6 Am^{-1}$
• D. $2.4 Am^{-1}$

Answer 1

The Correct Option is B

For a paramagnetic material, the magnetization \(M\) is related to the external magnetic field \(H\) and temperature \(T\) through Curie's Law, which is given by: \[ M \propto \frac{H}{T} \] Where:
\(M\) is the magnetization,
\(H\) is the external magnetic field,
\(T\) is the temperature in Kelvin.
Let the magnetization at the first condition (with \(H_1 = 0.6\) T and \(T_1 = 4\) K) be \(M_1 = 8 \, \text{A/m}\).
Now, the magnetization at the second condition (with \(H_2 = 0.2\) T and \(T_2 = 16\) K) can be calculated using the ratio: \[ \frac{M_1}{M_2} = \frac{H_1 / T_1}{H_2 / T_2} \] Substituting the values: \[ \frac{8}{M_2} = \frac{0.6 / 4}{0.2 / 16} \] Simplifying the ratio: \[ \frac{8}{M_2} = \frac{0.15}{0.0125} = 12 \] Thus, \[ M_2 = \frac{8}{12} = \frac{2}{3} \, \text{A/m} \]

Therefore, the magnetization at the second condition is \(\frac{2}{3}\) A/m, and the correct answer is (B).

Answer 2

For paramagnetic materials, the magnetization \( M \) is related to the external magnetic field \( H \) and temperature \( T \) by the Curie law: \[ M = C \frac{H}{T} \] where \( C \) is the Curie constant. From the given conditions:
At \( T = 4 \, \text{K} \) and \( H = 0.6 \, \text{T} \), the magnetization is \( M_1 = 8 \, \text{Am}^{-1} \).
At \( T = 16 \, \text{K} \) and \( H = 0.2 \, \text{T} \), we need to find the magnetization \( M_2 \).

Using the Curie law, we can write the ratio of the two magnetizations: \[ \frac{M_2}{M_1} = \frac{\frac{H_2}{T_2}}{\frac{H_1}{T_1}} = \frac{H_2 T_1}{H_1 T_2} \] Substitute the known values: \[ \frac{M_2}{8} = \frac{0.2 \times 4}{0.6 \times 16} \] \[ \frac{M_2}{8} = \frac{0.8}{9.6} = \frac{1}{12} \] \[ M_2 = \frac{8}{12} = \frac{2}{3} \, \text{Am}^{-1} \] Thus, the magnetization when the sample is placed in the second magnetic field at a temperature of 16 K is \( \frac{2}{3} \, \text{Am}^{-1} \).

Therefore, the correct answer is \({B} \).