A paramagnetic sample shows a net magnetisation of `8Am^-1` when placed in an external magnetic field of `0*6T` at a temperature of `4K`. When the same sample is placed in an external magnetic field of `0*2T` at a temperature of `16K`, the magnetisation will be

Options

A. `(32)/(3)Am^(-1)`
B. `(2)/(3)Am^(-1)`
C. `6Am^(-1)`
D. `2.4Am^(-1)

Explanation

To solve the problem, we will use Curie's law, which states that the magnetization \( M \) of a paramagnetic material is directly proportional to the magnetic field \( B \) and inversely proportional to the temperature \( T \). The relationship can be expressed as:

\[ M = \frac{C B}{T} \]

where \( C \) is a constant for the material.

### Step-by-step Solution:

1. **Identify Given Values:**
- For the first case:
- \( M_1 = 8 \, \text{A/m} \)
- \( B_1 = 0.6 \, \text{T} \)
- \( T_1 = 4 \, \text{K} \)
- For the second case:
- \( B_2 = 0.2 \, \text{T} \)
- \( T_2 = 16 \, \text{K} \)
- \( M_2 \) is what we need to find.

2. **Write the Equations Using Curie's Law:**
- For the first case:
\[ M_1 = \frac{C B_1}{T_1} \]
- For the second case:
\[ M_2 = \frac{C B_2}{T_2} \]

3. **Set Up the Ratio of Magnetizations:**
- From the two equations, we can set up a ratio:
\[ \frac{M_1}{M_2} = \frac{B_1}{B_2} \cdot \frac{T_2}{T_1} \]

4. **Substitute the Known Values:**
- Substitute \( M_1 = 8 \, \text{A/m} \), \( B_1 = 0.6 \, \text{T} \), \( B_2 = 0.2 \, \text{T} \), \( T_1 = 4 \, \text{K} \), and \( T_2 = 16 \, \text{K} \) into the ratio:
\[ \frac{8}{M_2} = \frac{0.6}{0.2} \cdot \frac{16}{4} \]

5. **Calculate the Right Side:**
- Calculate \( \frac{0.6}{0.2} = 3 \) and \( \frac{16}{4} = 4 \):
\[ \frac{8}{M_2} = 3 \cdot 4 = 12 \]

6. **Solve for \( M_2 \):**
- Rearranging gives:
\[ M_2 = \frac{8}{12} = \frac{2}{3} \, \text{A/m} \]

### Final Answer:
Thus, the magnetization \( M_2 \) when the sample is placed in an external magnetic field of \( 0.2 \, \text{T} \) at a temperature of \( 16 \, \text{K} \) is:
\[ M_2 = \frac{2}{3} \, \text{A/m} \]