A paramagnetic sample shows a net magnetisation of `0.8A-m^(-1)` when plced in an external mgnetic field of 0.8T at a temperature of 5K. Whent the same sample is placed in an external magnetic field of 0.4T at temperature of 20K, the magnetisation will be

Options

A. `0.8Am^(-1)`
B. `0.8Am^(-2)`
C. `0.1Am^(-1)`
D. `0.1Am^(-2)`

Explanation

To solve the problem, we will use the relationship between magnetization (I), magnetic field (B), and temperature (T) for paramagnetic materials, which can be expressed as:

\[ \frac{I_1}{I_2} = \frac{B_1}{B_2} \cdot \frac{T_2}{T_1} \]

Where:
- \(I_1\) is the initial magnetization (0.8 A/m)
- \(I_2\) is the magnetization we want to find
- \(B_1\) is the initial magnetic field (0.8 T)
- \(B_2\) is the new magnetic field (0.4 T)
- \(T_1\) is the initial temperature (5 K)
- \(T_2\) is the new temperature (20 K)

### Step-by-step Solution:

1. **Identify the given values**:
- \(I_1 = 0.8 \, \text{A/m}\)
- \(B_1 = 0.8 \, \text{T}\)
- \(B_2 = 0.4 \, \text{T}\)
- \(T_1 = 5 \, \text{K}\)
- \(T_2 = 20 \, \text{K}\)

2. **Set up the equation**:
Using the formula mentioned, we can write:
\[ \frac{0.8}{I_2} = \frac{0.8}{0.4} \cdot \frac{20}{5} \]

3. **Calculate the right side of the equation**:
- Calculate \(\frac{0.8}{0.4} = 2\)
- Calculate \(\frac{20}{5} = 4\)
- Therefore, the right side becomes:
\[ 2 \cdot 4 = 8 \]

4. **Substitute back into the equation**:
Now we have:
\[ \frac{0.8}{I_2} = 8 \]

5. **Solve for \(I_2\)**:
Rearranging gives:
\[ I_2 = \frac{0.8}{8} \]
\[ I_2 = 0.1 \, \text{A/m} \]

### Final Answer:
The magnetization \(I_2\) when the sample is placed in an external magnetic field of 0.4 T at a temperature of 20 K is \(0.1 \, \text{A/m}\).

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