A copper wire with a cross-section of 1.0 mm^2 carries a current of 1.0 A . Determine the drift speed of the free electrons in the wire, assuming there is one free electron per atom. The density of copper is 9000 kg m^

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Quantrex Academy · Accepted Answer

Correct answer: D. 0.074 mm s^ -1

The number density of free electrons is equal to the number density of copper atoms, given by:

n = \dfrac{N_A d}{M}

Substituting the standard values for copper (

N_A \approx 6.02 \times 10^{23} \text{ mol}^{-1}

,

M \approx 63.5 \text{ g mol}^{-1} = 63.5 \times 10^{-3} \text{ kg mol}^{-1}

):

n = \dfrac{6.02 \times 10^{23} \times 9000}{63.5 \times 10^{-3}} \approx 8.5 \times 10^{28} \text{ m}^{-3}

The drift speed

v_d

is given by the relation:

I = n e A v_d

Rearranging for

v_d

:

v_d = \dfrac{I}{n e A}

Substituting the given values (

I = 1.0 \text{ A}

,

e = 1.6 \times 10^{-19} \text{ C}

,

A = 1.0 \times 10^{-6} \text{ m}^2

):

v_d = \dfrac{1.0}{8.5 \times 10^{28} \times 1.6 \times 10^{-19} \times 1.0 \times 10^{-6}}

v_d \approx \dfrac{1.0}{13.6 \times 10^3} \text{ m s}^{-1} \approx 0.074 \times 10^{-3} \text{ m s}^{-1}

v_d = 0.074 \text{ mm s}^{-1}

A copper wire with a cross-section of 1.0 mm^2 carries a current of...

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