The correct interpretation of Ohm's law is: (Here, \( \vec{J} \) is current density, '\( \sigma \)' is conductivity, \( \vec{E} \) is electric field and '\( \rho \)' is resistivity.)
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → \( \vec{E}=\vec{J}\rho \)
The microscopic form of Ohm’s law says that the current density is proportional to the electric field, with the conductivity as the constant of proportionality:
\( \vec{J} = \sigma\vec{E} \)
Resistivity is the reciprocal of conductivity, \( \rho = 1/\sigma \). Dividing both sides by \( \sigma \):
\( \vec{E} = \frac{1}{\sigma}\,\vec{J} = \rho\,\vec{J} \)
Why the other options are incorrect
Option (1) \( \vec{E}=\sigma \vec{J} \) — Incorrect. It uses \( \sigma \) where \( \rho \) belongs; it would give \( \sigma \) the wrong dimensions.
Option (3) \( \vec{J}=\rho \vec{E} \) — Incorrect. It uses \( \rho \) where \( \sigma \) belongs, the reverse of the correct relation \( \vec{J}=\sigma\vec{E} \).
Option (4) \( \vec{J}=\frac{\sigma}{\vec{E}} \) — Incorrect. It makes J fall as E rises, but a larger field drives a larger current density.
NCERT (2026-27):
Ohm’s Law – “… The above relation for magnitudes E and j can indeed be cast in a vector form. … Thus, the last equation can be written as,
\( \displaystyle \mathbf{E} = \mathbf{j}\rho \qquad (3.12) \)
or,
\( \displaystyle \mathbf{j} = \sigma \mathbf{E} \qquad (3.13) \)
where σ ≡ 1/ρ is called the conductivity. Ohm’s law is often stated in an equivalent form, Eq. (3.13) in addition to Eq.(3.3).”