Drift Velocity
A copper wire of cross-sectional area (2 \times 10^{-6}) m(^2) carries a current of 3.2 A, with a free-electron density of (1.0 \times 10^{29}) m(^{-3}). What is the drift speed of the electrons?
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Solution
\(1.0 \times 10^{-4}\) m/s
Charge transport in a metal is described by relating the macroscopic current to the average velocity with which free electrons drift under an applied field, captured by (I = nAev_d), where (n) is the free-electron density, (A) the cross-section, (e) the electronic charge, and (v_d) the drift speed. Solving for the drift speed gives (v_d = I/(nAe)). Inserting (I = 3.2) A, (n = 1.0 \times 10^{29}) m(^{-3}), (A = 2 \times 10^{-6}) m(^2), and (e = 1.6 \times 10^{-19}) C gives (v_d = 3.2 / (1.0 \times 10^{29} \times 2 \times 10^{-6} \times 1.6 \times 10^{-19}) = 1.0 \times 10^{-4}) m/s. The value (1.0 \times 10^{-3}) m/s is wrong by a factor of ten, arising from a misplaced power of the area. The value (2.0 \times 10^{-4}) m/s mistakenly drops the factor of two in (A). The value (1.0 \times 10^{-5}) m/s overcounts the electron density. This follows the NCERT microscopic model of conduction. The remarkably small drift speed, despite a sizeable current, is the expected and well-known result, since the enormous electron number density compensates for the sluggish motion. It is worth emphasising that this drift speed is many orders of magnitude smaller than the random thermal speeds of the electrons, yet it is precisely this tiny field-driven net displacement, summed over a vast population of carriers, that constitutes the observable conduction current in the wire.
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About This Question
- Subject
- physics
- Chapter
- current electricity
- Topic
- drift velocity
- Difficulty
- Medium
- Year
- 2025
Solution
Correct Answer:
\(1.0 \times 10^{-4}\) m/s
Charge transport in a metal is described by relating the macroscopic current to the average velocity with which free electrons drift under an applied field, captured by (I = nAev_d), where (n) is the free-electron density, (A) the cross-section, (e) the electronic charge, and (v_d) the drift speed. Solving for the drift speed gives (v_d = I/(nAe)). Inserting (I = 3.2) A, (n = 1.0 \times 10^{29}) m(^{-3}), (A = 2 \times 10^{-6}) m(^2), and (e = 1.6 \times 10^{-19}) C gives (v_d = 3.2 / (1.0 \times 10^{29} \times 2 \times 10^{-6} \times 1.6 \times 10^{-19}) = 1.0 \times 10^{-4}) m/s. The value (1.0 \times 10^{-3}) m/s is wrong by a factor of ten, arising from a misplaced power of the area. The value (2.0 \times 10^{-4}) m/s mistakenly drops the factor of two in (A). The value (1.0 \times 10^{-5}) m/s overcounts the electron density. This follows the NCERT microscopic model of conduction. The remarkably small drift speed, despite a sizeable current, is the expected and well-known result, since the enormous electron number density compensates for the sluggish motion. It is worth emphasising that this drift speed is many orders of magnitude smaller than the random thermal speeds of the electrons, yet it is precisely this tiny field-driven net displacement, summed over a vast population of carriers, that constitutes the observable conduction current in the wire.
This medium difficulty physics question is from the chapter current electricity, covering the topic of drift velocity. It appeared in the 2025 exam.
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