First slide

Drift of electrons and the origin of resistivity

Question

Consider a wire of length 0.1 m with an area of cross section $1 {mm}^{2}$ connected to 5 V. Find the current flowing through the metallic wire where $μ = 5 \times 10^{- 6} m^{2} v^{- 1} s^{- 1}, e = 1.6 \times 10^{- 19} C$ and $n = 8 \times 10^{28} m^{- 3}$

Moderate

A

3.2 A
B

4.2 A
C

5.6 A
D

32 A

Solution

$E = \frac{V}{l} = \frac{5}{0.1} = 50 V m^{-1} \Rightarrow {I = neAV}_{d} = neAμE [∴ μ = V_{d} / E] = 8 \times 10^{28} \times 10^{- 6} \times 1.6 \times 10^{- 19} \times 5 \times 10^{- 6} \times 50 = 3.2 A$

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