A copper conductor of area of cross- section 40 `mm^(2)` on a side carries a constant current of ` 32 xx 10^(-6)A`. Then the current density is (in `amp//m^(2)`)

Options

A. 1.6
B. 0.8
C. 0.4
D. 3.2

Answer

To find the current density (J) in a copper conductor, we can use the formula:

\[ J = \frac{I}{A} \]

where:
- \( J \) is the current density in amperes per square meter (A/m²),
- \( I \) is the current in amperes (A),
- \( A \) is the cross-sectional area in square meters (m²).

### Step-by-Step Solution:

1. **Convert the Cross-Sectional Area from mm² to m²**:
- Given area \( A = 40 \, \text{mm}^2 \).
- To convert mm² to m², we use the conversion factor \( 1 \, \text{mm} = 10^{-3} \, \text{m} \).
- Therefore, \( A = 40 \, \text{mm}^2 = 40 \times (10^{-3} \, \text{m})^2 = 40 \times 10^{-6} \, \text{m}^2 \).

2. **Identify the Current**:
- Given current \( I = 32 \times 10^{-6} \, \text{A} \).

3. **Calculate the Current Density**:
- Using the formula for current density:
\[ J = \frac{I}{A} = \frac{32 \times 10^{-6} \, \text{A}}{40 \times 10^{-6} \, \text{m}^2} \]
- Simplifying this:
\[ J = \frac{32}{40} \, \text{A/m}^2 = 0.8 \, \text{A/m}^2 \]

4. **Final Result**:
- The current density \( J \) is \( 0.8 \, \text{A/m}^2 \).

### Conclusion:
The current density in the copper conductor is \( 0.8 \, \text{A/m}^2 \).