Exams › NEET › Physics

The electric field \(E\), current density \(J\), and conductivity \(\sigma\) of a conductor are related as

1. \(J = \sigma E\)
2. \(E = \sigma J\)
3. \(JE = \sigma\)
4. \(\frac{1}{JE} = \sigma\)

Correct answer: \(J = \sigma E\)

Solution

The microscopic form of Ohm's law is \(\vec{J} = \sigma \vec{E}\). Here, conductivity \(\sigma\) measures how easily current flows in response to an electric field. So the correct relation is \(J = \sigma E\).

Related NEET Physics questions

• For a metallic wire, the ratio (V)/(i) (where, V= applied potential difference and i= current flowing
• In a metallic conductor, electric current is thought to be due to the movement of
• Assertion If the length of the conductor is doubled, the drift velocity will become half of the original value (keeping potential difference unchanged). Reason At constant potential difference, drift velocity is inversely proportional to the length of the conductor.
• The resistance of an iron wire is 10 $\Omega$ and its temperature coefficient of resistance is $5 \times 10^{-3}\,/^\circ\text{C}$. A current of 30 mA is flowing in it at $20^\circ\text{C}$. Keeping the potential difference across its ends constant, if its temperature is increased to $120^\circ\text{C}$, then the current flowing in the wire will be (in mA):
• A steady current of magnitude \(I\) and an AC current of peak value \(I\) are passed through identical resistors for the same time. The ratio of heat produced in the two resistors is:
• Wire of a certain material is stretched slowly be dep10. Its new resistance and specific resistance becomes respectively.

⚔️ Practice NEET Physics free + battle 1v1 →