A metal sample carrying a current along x axis with density is subjected to a magnetic field `B_(z)` (along z - axis). The electric field `E_(y)` developed along Y-axis is directly proportional to `J_(x)` as well as `B_(z).` The constant of proportionality has SI unit.

Options

A. `(m^(2))/(A)`
B. `(m^(3))/(As)`
C. `(m^(2))/(As)`
D. `(As)/(m^(3))`

Explanation

To solve the problem, we need to analyze the relationship between the electric field \( E_y \), the current density \( J_x \), and the magnetic field \( B_z \).

### Step-by-Step Solution:

1. **Understanding the Relationship**:
The problem states that the electric field \( E_y \) developed along the y-axis is directly proportional to the current density \( J_x \) (which flows along the x-axis) and the magnetic field \( B_z \) (which is along the z-axis). This can be expressed mathematically as:
\[ E_y \propto J_x \cdot B_z \]

2. **Introducing the Proportionality Constant**:
We can introduce a constant of proportionality \( K \) such that:
\[ E_y = K \cdot J_x \cdot B_z \]

3. **Identifying the Units**:
We need to determine the SI units of the constant \( K \). The units of \( E_y \), \( J_x \), and \( B_z \) are as follows:
- The unit of electric field \( E_y \) is volts per meter (V/m).
- The unit of current density \( J_x \) is amperes per square meter (A/m²).
- The unit of magnetic field \( B_z \) is teslas (T), which can also be expressed as kg/(A·s²).

4. **Expressing Units in Terms of Base SI Units**:
We can express these units in terms of base SI units:
- \( [E_y] = \text{V/m} = \text{(kg·m²)/(s³·A)} \)
- \( [J_x] = \text{A/m²} \)
- \( [B_z] = \text{T} = \text{kg/(A·s²)} \)

5. **Substituting the Units into the Equation**:
Now substituting the units into the equation \( E_y = K \cdot J_x \cdot B_z \):
\[ [E_y] = [K] \cdot [J_x] \cdot [B_z] \]
\[ \text{(kg·m²)/(s³·A)} = [K] \cdot \text{(A/m²)} \cdot \text{(kg/(A·s²))} \]

6. **Simplifying the Units**:
Rearranging the right-hand side:
\[ [K] \cdot \left(\frac{\text{A}}{\text{m²}}\right) \cdot \left(\frac{\text{kg}}{\text{A·s²}}\right) = [K] \cdot \frac{\text{kg}}{\text{m²·s²}} \]
Therefore, we have:
\[ \text{(kg·m²)/(s³·A)} = [K] \cdot \frac{\text{kg}}{\text{m²·s²}} \]

7. **Solving for \( K \)**:
To isolate \( K \), we can divide both sides by \( \frac{\text{kg}}{\text{m²·s²}} \):
\[ [K] = \frac{\text{(kg·m²)/(s³·A)}}{\text{kg/(m²·s²)}} = \frac{m²·s²}{s³·A} = \frac{m²}{s·A} \]

8. **Final Result**:
Thus, the unit of the constant of proportionality \( K \) is:
\[ [K] = \text{m²/(s·A)} \]