The ratio of electric field and current density in a current carrying conductor is called
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → Resistivity
Ohm’s law in vector form links the current density to the field in the conductor:
\( \vec{J} = \sigma\vec{E} \quad \text{or} \quad \vec{E} = \rho\vec{J} \)
So the ratio \( E/J \) is the resistivity \( \rho \), and its reciprocal \( J/E \) is the conductivity \( \sigma \).
Why the other options are incorrect
Option (1) Mobility — Incorrect. Mobility is drift speed per unit field, \( v_{d}/E \), not \( E/J \).
Option (2) Resistance — Incorrect. Resistance is \( V/I \). It depends on the length and area of the conductor, while \( E/J \) is a property of the material alone.
Option (3) Conductivity — Incorrect. Conductivity is the inverse ratio, \( J/E \).
NCERT (2026-27):
Ohm’s Law – “… 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 \( \sigma \equiv 1/\rho \) is called the conductivity. Ohm’s law is often stated in an equivalent form, Eq. (3.13) in addition to Eq.(3.3).”