Question

If \( I = 16 \, \text{A} \), electron density \( n = 5 \times 10^{23} \, \text{m}^{-3} \), and \( A = 1 \times 10^{-7} \, \text{m}^2 \), find the drift velocity.

Hint:

When calculating the drift velocity, use the formula \( I = n e A v_d \) and rearrange it to solve for \( v_d \).

Answer 1

The current \( I \) is related to the drift velocity \( v_d \) by the equation: \[ I = n e A v_d \] where:
- \( I \) is the current,
- \( n \) is the electron density,
- \( e \) is the charge of an electron (\( e = 1.6 \times 10^{-19} \, \text{C} \)),
- \( A \) is the cross-sectional area of the conductor, - \( v_d \) is the drift velocity.
Step 1: Rearrange the equation to solve for \( v_d \).
We can rearrange the equation to solve for \( v_d \): \[ v_d = \frac{I}{n e A} \]
Step 2: Substitute the given values.
Substitute the given values into the equation: \[ v_d = \frac{16}{(5 \times 10^{23}) \times (1.6 \times 10^{-19}) \times (1 \times 10^{-7})} \]
Step 3: Calculate the drift velocity.
Now, calculate the value of \( v_d \): \[ v_d = \frac{16}{(5 \times 10^{23}) \times (1.6 \times 10^{-19}) \times (1 \times 10^{-7})} = \frac{16}{8 \times 10^{7}} = 2 \times 10^{-7} \, \text{m/s} \]
Step 4: Conclusion.
The drift velocity is: \[ \boxed{2 \times 10^{-7} \, \text{m/s}} \]