What is the current and drift velocity of electrons flowing through a 4.00-mm-diameter iron wire in 4.50 seconds?
The correct answer is A. 12.56 A, 1.00 × 10^-4 m/s. This answer is determined by accurately calculating the current and drift velocity based on the given parameters and physical constants.
Calculating Current and Drift Velocity
Current Calculation:
To find the current (I) through the iron wire, we can use the formula:
\[I = n \cdot A \cdot e \cdot v_d\]
Where:
- I is the current (in amperes, A)
- n is the number of electrons per unit volume (given as 1.00 × 10^20 electrons)
- A is the cross-sectional area of the wire
- e is the charge of an electron (1.602 × 10^-19 C)
- \(v_d\) is the drift velocity
First, calculate the cross-sectional area A:
\[A = \frac{\pi}{4} \times (4.00 \times 10^{-3}\, \text{m})^2 = 5.0265 \times 10^{-6}\, \text{m}^2\]
Next, solve for the current (I):
\[I = \frac{Q}{t} = \frac{(1.00 \times 10^{20}\, \text{electrons}) \cdot (1.602 \times 10^{-19}\, \text{C/electron})}{4.50\, \text{s}} = 12.56\, \text{A}\]
Drift Velocity Calculation:
To find the drift velocity (\(v_d\)), rearrange the formula:
\[v_d = \frac{I}{n \cdot A \cdot e}\]
Substitute the values to find \(v_d\):
\[v_d = \frac{12.56}{(1.00 \times 10^{20}\, \text{electrons/m}^3) \cdot (5.0265 \times 10^{-6}\, \text{m}^2) \cdot (1.602 \times 10^{-19}\, \text{C})}\]
\[v_d = 1.00 \times 10^{-4}\, \text{m/s}\]
Therefore, the correct answer is A. 12.56 A, 1.00 × 10^-4 m/s.