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

Options

A. `(32)/(3) Am^(-1)`
B. `2/3 Am^(-1)`
C. `6Am^(-1)`
D. `2.4 Am^(-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 applied magnetic field \( B \) and inversely proportional to the absolute temperature \( T \). The relationship can be expressed as:

\[ M \propto \frac{B}{T} \]

### Step-by-Step Solution:

1. **Identify Given Values:**
- For the first condition:
- Magnetization \( M_1 = 8 \, \text{A/m} \)
- Magnetic field \( B_1 = 0.6 \, \text{T} \)
- Temperature \( T_1 = 4 \, \text{K} \)

- For the second condition:
- Magnetic field \( B_2 = 0.2 \, \text{T} \)
- Temperature \( T_2 = 16 \, \text{K} \)

2. **Apply Curie’s Law:**
According to Curie’s Law, we can write the ratio of magnetizations for the two conditions as follows:

\[ \frac{M_2}{M_1} = \frac{B_2 / T_2}{B_1 / T_1} \]

3. **Substitute the Known Values:**
Plugging in the known values into the equation:

\[ \frac{M_2}{8} = \frac{0.2 / 16}{0.6 / 4} \]

4. **Simplify the Right Side:**
- Calculate the right side:
\[ \frac{0.2 / 16}{0.6 / 4} = \frac{0.2 \times 4}{0.6 \times 16} = \frac{0.8}{9.6} = \frac{1}{12} \]

5. **Solve for \( M_2 \):**
Now we can solve for \( M_2 \):
\[ M_2 = 8 \times \frac{1}{12} = \frac{8}{12} = \frac{2}{3} \, \text{A/m} \]

### Final Answer:
The magnetization \( M_2 \) when the sample is placed in an external magnetic field of 0.2 T at a temperature of 16 K is:

\[ M_2 = \frac{2}{3} \, \text{A/m} \]