Question

\(\overrightarrow{E}\) is the electric field inside a conductor whose material has conductivity σ and resistivity ρ . The current density inside the conductor is \(\overrightarrow{j}\) . The correct form of Ohm's law is

• A. \(\overrightarrow{E}=\sigma\overrightarrow{j}\)
• B. \(\overrightarrow{j}=\rho\overrightarrow{E}\)
• C. \(\overrightarrow{E}=\rho\overrightarrow{j}\)
• D. \(\overrightarrow{E}. \overrightarrow{j}=\rho\)

Answer 1

The Correct Option is C

Ohm's law in its microscopic form relates the electric field $\vec{E}$ in a conductor to the current density $\vec{j}$ by: \[ \vec{j} = \sigma \vec{E} \] Here, $\sigma$ is the conductivity of the material. To find the electric field in terms of current density, rearrange the equation: \[ \vec{E} = \frac{1}{\sigma} \vec{j} \] Since resistivity $\rho = \frac{1}{\sigma}$, we substitute: \[ \vec{E} = \rho \vec{j} \] Thus, the correct form of Ohm's law in terms of resistivity is $\vec{E} = \rho \vec{j}$.