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Math  /  Algebra

Question1. Calculate the average drift velocity of electrons when a current of 1.5 A is passing through a silver wire of 1.0 mm21.0 \mathrm{~mm}^{2} cross-sectional area

Studdy Solution

STEP 1

Assumptions
1. The current passing through the wire is I=1.5 A I = 1.5 \, \text{A} .
2. The cross-sectional area of the wire is A=1.0 mm2=1.0×10−6 m2 A = 1.0 \, \text{mm}^2 = 1.0 \times 10^{-6} \, \text{m}^2 .
3. The charge of an electron is e=1.6×10−19 C e = 1.6 \times 10^{-19} \, \text{C} .
4. The density of conduction electrons in silver is approximately n=5.8×1028 electrons/m3 n = 5.8 \times 10^{28} \, \text{electrons/m}^3 .

STEP 2

The formula for the drift velocity vd v_d of electrons is given by:
vd=In⋅A⋅e v_d = \frac{I}{n \cdot A \cdot e}
where: - I I is the current, - n n is the electron density, - A A is the cross-sectional area, - e e is the charge of an electron.

STEP 3

Substitute the known values into the formula:
vd=1.55.8×1028×1.0×10−6×1.6×10−19 v_d = \frac{1.5}{5.8 \times 10^{28} \times 1.0 \times 10^{-6} \times 1.6 \times 10^{-19}}

STEP 4

Calculate the denominator:
5.8×1028×1.0×10−6×1.6×10−19=9.28×103 5.8 \times 10^{28} \times 1.0 \times 10^{-6} \times 1.6 \times 10^{-19} = 9.28 \times 10^{3}

STEP 5

Calculate the drift velocity vd v_d :
vd=1.59.28×103 v_d = \frac{1.5}{9.28 \times 10^{3}}
vd≈1.617×10−4 m/s v_d \approx 1.617 \times 10^{-4} \, \text{m/s}
The average drift velocity of electrons is approximately 1.617×10−4 m/s 1.617 \times 10^{-4} \, \text{m/s} .

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