A metal sample carrying a current along X-axis with density $$J_x$$ 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.

We are given that the electric field $$E_y$$ along the Y-axis is directly proportional to both the current density $$J_x$$ along the X-axis and the magnetic field $$B_z$$ along the Z-axis. This relationship can be expressed as:

$$ E_y = R \cdot J_x \cdot B_z $$

where $$R$$ is the constant of proportionality whose SI unit we need to find.

To determine the unit of $$R$$, we first recall the SI units of each quantity:

• The electric field $$E_y$$ has the SI unit volt per meter (V/m). Since 1 volt = 1 joule per coulomb (J/C) and 1 joule = 1 newton meter (N·m), we can write: $$ \text{V} = \frac{\text{J}}{\text{C}} = \frac{\text{N} \cdot \text{m}}{\text{C}} $$ Therefore, $$ \text{V/m} = \frac{\text{N} \cdot \text{m}}{\text{C} \cdot \text{m}} = \frac{\text{N}}{\text{C}} $$ Now, force (newton, N) has base units kg·m·s⁻², and charge (coulomb, C) has base units ampere second (A·s). Substituting: $$ \text{V/m} = \frac{\text{kg} \cdot \text{m} \cdot \text{s}^{-2}}{\text{A} \cdot \text{s}} = \text{kg} \cdot \text{m} \cdot \text{s}^{-3} \cdot \text{A}^{-1} $$
• The current density $$J_x$$ is defined as current per unit area, so its SI unit is ampere per square meter (A/m²).
• The magnetic field $$B_z$$ has the SI unit tesla (T). Since 1 tesla = 1 weber per square meter (Wb/m²) and 1 weber = 1 volt second (V·s), we have: $$ \text{T} = \frac{\text{V} \cdot \text{s}}{\text{m}^2} $$ Substituting the base units for volt (V = kg·m·s⁻³·A⁻¹): $$ \text{T} = \frac{(\text{kg} \cdot \text{m} \cdot \text{s}^{-3} \cdot \text{A}^{-1}) \cdot \text{s}}{\text{m}^2} = \frac{\text{kg} \cdot \text{m} \cdot \text{s}^{-2} \cdot \text{A}^{-1}}{\text{m}^2} = \text{kg} \cdot \text{s}^{-2} \cdot \text{A}^{-1} $$

From the equation $$E_y = R \cdot J_x \cdot B_z$$, we solve for $$R$$:

$$ R = \frac{E_y}{J_x \cdot B_z} $$

The unit of $$R$$ is the unit of $$E_y$$ divided by the product of the units of $$J_x$$ and $$B_z$$. Substituting the units:

$$ \text{Unit of } R = \frac{\text{kg} \cdot \text{m} \cdot \text{s}^{-3} \cdot \text{A}^{-1}}{(\text{A}/\text{m}^2) \cdot (\text{kg} \cdot \text{s}^{-2} \cdot \text{A}^{-1})} $$

First, simplify the denominator:

$$ (\text{A}/\text{m}^2) \cdot (\text{kg} \cdot \text{s}^{-2} \cdot \text{A}^{-1}) = \frac{\text{A}}{\text{m}^2} \cdot \frac{\text{kg}}{\text{s}^2 \cdot \text{A}} = \frac{\text{kg}}{\text{m}^2 \cdot \text{s}^2} $$

Now substitute back:

$$ \text{Unit of } R = \frac{\text{kg} \cdot \text{m} \cdot \text{s}^{-3} \cdot \text{A}^{-1}}{\frac{\text{kg}}{\text{m}^2 \cdot \text{s}^2}} = \text{kg} \cdot \text{m} \cdot \text{s}^{-3} \cdot \text{A}^{-1} \times \frac{\text{m}^2 \cdot \text{s}^2}{\text{kg}} $$

Cancel common units:

• kg in numerator and denominator cancel.
• s⁻³ and s² combine to s⁻¹ (since s⁻³ × s² = s⁻¹).
• m and m² combine to m³.
• A⁻¹ remains.

So we get:

$$ \text{Unit of } R = \text{m}^3 \cdot \text{s}^{-1} \cdot \text{A}^{-1} = \frac{\text{m}^3}{\text{A} \cdot \text{s}} $$

Now, comparing with the options:

A. $$\frac{m^2}{A}$$

B. $$\frac{m^3}{As}$$

C. $$\frac{m^2}{As}$$

D. $$\frac{As}{m^3}$$

Option B matches $$\frac{m^3}{As}$$, where "As" denotes ampere second (A·s).

Hence, the correct answer is Option B.