Question:
An electron takes \(40\times10^3\;s\) to drift from one end of a metal wire of length \(2\;m\) to its other end. The area of cross-section of the wire is \(4\;mm^2\) and it is carrying a current of \(1.6\;A\). The number density of free electrons in the metal wire is
An electron takes \(40\times10^3\;s\) to drift from one end of a metal wire of length \(2\;m\) to its other end. The area of cross-section of the wire is \(4\;mm^2\) and it is carrying a current of \(1.6\;A\). The number density of free electrons in the metal wire is
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For current through a conductor, use
\[
I=neAv_d
\]
where \(n\) is number density, \(e\) is electronic charge, \(A\) is area, and \(v_d\) is drift velocity.
Updated On: Jun 22, 2026
- \(8\times10^{28}\;m^{-3}\)
- \(6\times10^{28}\;m^{-3}\)
- \(4\times10^{28}\;m^{-3}\)
- \(5\times10^{28}\;m^{-3}\)
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The Correct Option is D
Solution and Explanation
Step 1: Find drift velocity.
Length of wire is \[ L=2\;m \] Time taken by electron is \[ t=40\times10^3\;s \] Drift velocity is \[ v_d=\frac{L}{t} \] \[ v_d=\frac{2}{40\times10^3} \] \[ v_d=5\times10^{-5}\;m s^{-1} \]
Step 2: Convert area into SI unit.
\[ A=4\;mm^2 \] \[ A=4\times10^{-6}\;m^2 \]
Step 3: Use current formula.
Current is given by \[ I=neAv_d \] So, \[ n=\frac{I}{eAv_d} \] Given, \[ I=1.6\;A \] and \[ e=1.6\times10^{-19}\;C \]
Step 4: Substitute the values.
\[ n=\frac{1.6}{(1.6\times10^{-19})(4\times10^{-6})(5\times10^{-5})} \] \[ n=\frac{1.6}{3.2\times10^{-29}} \] \[ n=5\times10^{28}\;m^{-3} \]
Step 5: Final conclusion.
Hence, the number density of free electrons is \[ \boxed{5\times10^{28}\;m^{-3}} \]
Length of wire is \[ L=2\;m \] Time taken by electron is \[ t=40\times10^3\;s \] Drift velocity is \[ v_d=\frac{L}{t} \] \[ v_d=\frac{2}{40\times10^3} \] \[ v_d=5\times10^{-5}\;m s^{-1} \]
Step 2: Convert area into SI unit.
\[ A=4\;mm^2 \] \[ A=4\times10^{-6}\;m^2 \]
Step 3: Use current formula.
Current is given by \[ I=neAv_d \] So, \[ n=\frac{I}{eAv_d} \] Given, \[ I=1.6\;A \] and \[ e=1.6\times10^{-19}\;C \]
Step 4: Substitute the values.
\[ n=\frac{1.6}{(1.6\times10^{-19})(4\times10^{-6})(5\times10^{-5})} \] \[ n=\frac{1.6}{3.2\times10^{-29}} \] \[ n=5\times10^{28}\;m^{-3} \]
Step 5: Final conclusion.
Hence, the number density of free electrons is \[ \boxed{5\times10^{28}\;m^{-3}} \]
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