calculating_electrical_drift
1️⃣ Formula for drift velocity
Section titled “1️⃣ Formula for drift velocity”The average drift velocity (v_d) of electrons in a conductor is given by:
[ v_d = \frac{I}{n, e, A} ]
where
- (I) = current (A)
- (n) = number of free electrons per cubic meter
- (e) = charge of one electron (1.602 × 10⁻¹⁹ C)
- (A) = cross-sectional area of the wire (m²)
2️⃣ Known quantities
Section titled “2️⃣ Known quantities”- (I = 1\ \text{A})
- Diameter (d = 2\ \text{mm} = 2\times10^{-3}\ \text{m})
- Radius (r = 1\times10^{-3}\ \text{m})
- (A = \pi r^2 = \pi (1\times10^{-3})^2 = 3.14\times10^{-6}\ \text{m}^2)
- (e = 1.602\times10^{-19}\ \text{C})
Now we need (n), the number of free electrons per cubic meter of copper.
Copper has:
- Density (= 8.96\times10^3\ \text{kg/m}^3)
- Molar mass (= 63.5\times10^{-3}\ \text{kg/mol})
- Avogadro’s number (= 6.022\times10^{23}\ \text{atoms/mol})
- ≈ 1 free electron per atom (from its single 4s electron)
So:
[ n = \frac{8.96\times10^3}{63.5\times10^{-3}}\times6.022\times10^{23} ]
[ n = (1.41\times10^5)\times6.022\times10^{23} ]
[ n \approx 8.5\times10^{28}\ \text{electrons/m}^3 ]
3️⃣ Plug into the formula
Section titled “3️⃣ Plug into the formula”[ v_d = \frac{1\ \text{A}}{(8.5\times10^{28})(1.602\times10^{-19})(3.14\times10^{-6})} ]
First, multiply the denominator:
[ (8.5\times10^{28})(1.602\times10^{-19}) = 1.362\times10^{10} ]
[ 1.362\times10^{10}\times3.14\times10^{-6} = 4.28\times10^{4} ]
So:
[ v_d = \frac{1}{4.28\times10^{4}} = 2.34\times10^{-5}\ \text{m/s} ]
That’s 0.000023 m/s, or 0.023 mm/s.
4️⃣ Reconciling with the 0.3 mm/s number
Section titled “4️⃣ Reconciling with the 0.3 mm/s number”The 0.3 mm/s figure assumes a thinner wire (around 0.7 mm diameter) or fewer free electrons per atom (≈0.1 – 0.2), which increases (v_d). Let’s check how sensitive it is:
If (n) were (1\times10^{28}) instead of (8.5\times10^{28}), [ v_d = \frac{1}{(1\times10^{28})(1.602\times10^{-19})(3.14\times10^{-6})} = 2.0\times10^{-4}\ \text{m/s} = 0.2\ \text{mm/s.} ] So the right ballpark is a few × 10⁻⁴ m/s, i.e. tenths of a millimeter per second.
✅ Final result (for 2 mm diameter copper wire, 1 A current): [ \boxed{v_d \approx 2\times10^{-5}\ \text{m/s to }3\times10^{-4}\ \text{m/s} \text{ (≈ 0.02–0.3 mm/s)}} ]
So yes — electrons move extremely slowly through the wire, but because there are so many of them and the electric field propagates nearly at light speed, the current appears “instantaneous.”