A wire with an area of cross-section as $10 m m^{2}$ has a resistance of $5 Ω$ , when a potential difference across its ends is $25 V$ . Calculate the drift velocity of electrons. Given the number density of electrons as $5 \times 10^{20}$ electrons per cubic meter $(e / m^{- 3})$ .

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Solution

Given : $A = 10 m m^{2} = 10 \times 10^{- 6} m^{2}$ $n = 5 \times 10^{20} / m^{3}$
Current $I = \frac{V}{R} = \frac{25}{5} = 5 A$
Drift velocity $V_{d} = \frac{I}{A n e}$
$⟹ V_{d} = \frac{5}{10 \times 10^{- 6} \times 5 \times 10^{20} \times 1.6 \times 10^{- 19}} = 6.25 \times 10^{3} m / s$

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A wire with an area of cross-section as 10mm2 has a resistance of 5Ω, when a potential difference across its ends is 25V. Calculate the drift velocity of electrons. Given the number density of electrons as 5×1020 electrons per cubic meter (e/m−3).

Given : A=10 mm2=10×10−6 m2 n=5×1020 /m3 Current I=VR=255=5 ADrift velocity Vd=IAne⟹ Vd=510×10−6×5×1020×1.6×10−19=6.25×103 m/s

A wire with an area of cross-section as $10 m m^{2}$ has a resistance of $5 Ω$ , when a potential difference across its ends is $25 V$ . Calculate the drift velocity of electrons. Given the number density of electrons as $5 \times 10^{20}$ electrons per cubic meter $(e / m^{- 3})$ .

Given : $A = 10 m m^{2} = 10 \times 10^{- 6} m^{2}$ $n = 5 \times 10^{20} / m^{3}$
Current $I = \frac{V}{R} = \frac{25}{5} = 5 A$
Drift velocity $V_{d} = \frac{I}{A n e}$
$⟹ V_{d} = \frac{5}{10 \times 10^{- 6} \times 5 \times 10^{20} \times 1.6 \times 10^{- 19}} = 6.25 \times 10^{3} m / s$