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Difficulty: MediumElectric Current and Resistance

A conductor wire with a cross-sectional area of 2.0×10−6 m22.0 \times 10^{-6}\,\text{m}^2 carries a steady current of 1.6 A1.6\,\text{A}. If the conduction electron density of the material is 5.0×1028 m−35.0 \times 10^{28}\,\text{m}^{-3} and the elementary charge is 1.6×10−19 C1.6 \times 10^{-19}\,\text{C}, what is the average drift velocity of the electrons in the wire?

  1. 1.0×10−4 m/s1.0 \times 10^{-4}\,\text{m/s}Answer
  2. B
    1.0×10−6 m/s1.0 \times 10^{-6}\,\text{m/s}
  3. C
    1.6×10−23 m/s1.6 \times 10^{-23}\,\text{m/s}
  4. D
    4.0×10−16 m/s4.0 \times 10^{-16}\,\text{m/s}

Answer

The average drift velocity of the electrons is 1.0×10−4 m/s1.0 \times 10^{-4}\,\text{m/s}.
The correct answer is derived from the fundamental relationship I=nAevdI = n A e v_d. Solving for drift velocity gives vd=InAe=1.65.0×1028×2.0×10−6×1.6×10−19=1.0×10−4 m/sv_d = \frac{I}{n A e} = \frac{1.6}{5.0 \times 10^{28} \times 2.0 \times 10^{-6} \times 1.6 \times 10^{-19}} = 1.0 \times 10^{-4}\,\text{m/s}.

Step-by-Step Solution

1
Identify the drift velocity formula relating current to charge carrier parameters
The electric current is given by I=nAevdI = n A e v_d, where II is current, nn is electron density, AA is cross-sectional area, ee is elementary charge, and vdv_d is drift velocity.
This formula connects macroscopic electric current to microscopic charge dynamics.
2
Rearrange the equation to solve for drift velocity vdv_d
vd=InAev_d = \frac{I}{n A e}
Isolating the unknown variable vdv_d before substituting known values.
3
Substitute the given values into the expression
vd=1.6(5.0×1028)×(2.0×10−6)×(1.6×10−19)v_d = \frac{1.6}{(5.0 \times 10^{28}) \times (2.0 \times 10^{-6}) \times (1.6 \times 10^{-19})}
Inserting I=1.6 AI = 1.6\,\text{A}, n=5.0×1028 m−3n = 5.0 \times 10^{28}\,\text{m}^{-3}, A=2.0×10−6 m2A = 2.0 \times 10^{-6}\,\text{m}^2, and e=1.6×10−19 Ce = 1.6 \times 10^{-19}\,\text{C}.
4
Evaluate the denominator and compute the final value of vdv_d
Denominator =(5.0×2.0×1.6)×1028−6−19=16.0×103=1.6×104= (5.0 \times 2.0 \times 1.6) \times 10^{28 - 6 - 19} = 16.0 \times 10^3 = 1.6 \times 10^4. Thus, vd=1.61.6×104=1.0×10−4 m/sv_d = \frac{1.6}{1.6 \times 10^4} = 1.0 \times 10^{-4}\,\text{m/s}.
Simplifying powers of ten gives the final numerical answer.

Key Concept

Relationship between Electric Current and Drift Velocity
Estimated Time:1m 30s
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Topics

Electric Current and Resistance

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