QuestionJune 8, 2025

The density of free electrons in gold is 5.90times 10^28electrons/m^3 Find the drift speed of electrons in a gold wire of diameter 2.98 mm when a current of 0.175 A flows through the wire. 17.8times 10^-6m/s 4.32times 10^-6m/s 15.8times 10^-8m/s 2.65times 10^-6m/s 3.00times 10^8m/s

The density of free electrons in gold is 5.90times 10^28electrons/m^3 Find the drift speed of electrons in a gold wire of diameter 2.98 mm when a current of 0.175 A flows through the wire. 17.8times 10^-6m/s 4.32times 10^-6m/s 15.8times 10^-8m/s 2.65times 10^-6m/s 3.00times 10^8m/s
The density of free electrons in gold is 5.90times 10^28electrons/m^3 Find the drift speed of electrons in a gold wire of diameter 2.98 mm when a current of 0.175 A flows through the wire. 17.8times 10^-6m/s 4.32times 10^-6m/s 15.8times 10^-8m/s 2.65times 10^-6m/s 3.00times 10^8m/s

Solution
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Answer

4.32\times 10^{-6}m/s Explanation 1. Calculate the cross-sectional area Diameter = 2.98 mm = 0.00298 m. Area A = \pi \left(\frac{d}{2}\right)^2 = \pi \left(\frac{0.00298}{2}\right)^2. 2. Use formula for drift speed **Drift speed** v_d = \frac{I}{n \cdot e \cdot A}, where I = 0.175 A, n = 5.90 \times 10^{28} \text{ electrons/m}^3, e = 1.60 \times 10^{-19} \text{ C}. 3. Substitute values and calculate Substitute A, I, n, and e into the formula to find v_d.

Explanation

1. Calculate the cross-sectional area<br /> Diameter = 2.98 mm = 0.00298 m. Area $A = \pi \left(\frac{d}{2}\right)^2 = \pi \left(\frac{0.00298}{2}\right)^2$.<br />2. Use formula for drift speed<br /> **Drift speed** $v_d = \frac{I}{n \cdot e \cdot A}$, where $I = 0.175$ A, $n = 5.90 \times 10^{28} \text{ electrons/m}^3$, $e = 1.60 \times 10^{-19} \text{ C}$.<br />3. Substitute values and calculate<br /> Substitute $A$, $I$, $n$, and $e$ into the formula to find $v_d$.
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