Question:medium

A conducting wire of length L, uniform area of cross-section A, and material having n free electrons per unit volume offers a resistance R to the flow of current. m and e are the mass and charge of an electron, respectively. If τ is the mean free time of the electrons in the conductor, the correct formula for resistance R is:

Updated On: Mar 27, 2026
  • \(\quad R = \frac{mL}{e^2 n A \tau} \\\)
  • \(\quad R = \frac{mA}{e^2 n L \tau} \\\)
  • \(\quad R = \frac{m \tau}{e^2 n A L} \\\)
  • \(\quad R = \frac{e^2 n A \tau}{m L}\)
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The Correct Option is A

Solution and Explanation

A conducting wire possesses length \(L\), uniform cross-sectional area \(A\), and is composed of a material with \(n\) free electrons per unit volume. Its resistance \(R\) is determined by these parameters, along with the electron mass \(m\), electron charge \(e\), and the electrons' mean free time \(\tau\). This document aims to derive the formula for \(R\).

1. Drift Velocity and Current Density: Current density \(J\) is defined as:

\[J=n \cdot e \cdot v_d\]

where \(v_d\) represents the drift velocity. Ohm's Law states:

\[J=\sigma \cdot E\]

where \(\sigma\) is the material's conductivity and \(E\) is the electric field.

2. Conductivity and Relaxation Time: The relationship between conductivity and relaxation time \(\tau\) is:

\[\sigma=\frac{n e^2 \tau}{m}\]

Consequently, Ohm's Law can be written as:

\[J=\frac{n e^2 \tau}{m} \cdot E\]

Equating the two expressions for current density yields:

\[n e v_d=\frac{n e^2 \tau}{m} \cdot E\]

3. Electric Field and Voltage: For a conductor of length \(L\), the electric field \(E\) is related to the voltage \(V\) by:

\[E=\frac{V}{L}\]

Substituting this relationship, we obtain:

\[v_d=\frac{e \tau}{m} \cdot \frac{V}{L}\]

4. Resistivity to Resistance Conversion: Resistivity \(\rho\) is the inverse of conductivity:

\[\rho=\frac{m}{n e^2 \tau}\]

Resistance \(R\) is calculated as:

\[R=\rho \cdot \frac{L}{A}\]

Substituting the expression for \(\rho\):

\[R=\frac{m}{n e^2 \tau} \cdot \frac{L}{A}\]

Simplification leads to:

\[R=\frac{mL}{e^2 n A \tau}\]

5. Conclusion: The derived formula for resistance \(R\) is:

\[R=\frac{mL}{e^2 n A \tau}\]

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