A wire of length 4 m and cross-sectional area 1 mm^2 carries a current of 2 A . Given that the material holds 10^ 29 free electrons per cubic metre, determine the average time required for an electron to traverse the

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Quantrex Academy · Accepted Answer

Correct answer: D. 3.2 × 10^4 s

The drift velocity of free electrons is given by the relation

I = n e A v_d

, which gives

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

.

The average time required for an electron to traverse the length

L

of the wire is

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

.

Substituting the given values:

L = 4 \text{ m}

n = 10^{29} \text{ m}^{-3}

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

A = 1 \text{ mm}^2 = 10^{-6} \text{ m}^2

I = 2 \text{ A}

t = \dfrac{4 \times 10^{29} \times 1.6 \times 10^{-19} \times 10^{-6}}{2}

t = \dfrac{6.4 \times 10^4}{2}

t = 3.2 \times 10^4 \text{ s}

A wire of length 4 m and cross-sectional area 1 mm^2 carries a...

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