Question

The number density of free electrons in copper is nearly 8 × 10²⁸ m^-3. A copper wire has its area of cross section = 2 × 10^-6 m² and is carrying a current of 3.2 A. The drift speed of the electrons is _____ ×10^-6 ms^-1

Answer 1

Correct Answer: 125

The drift speed \( v_d \) of electrons in a conductor can be determined using the formula for current, \( I = nAev_d \), where:

• \( I \) is the current (3.2 A)
• \( n \) is the number density of free electrons (\( 8 \times 10^{28} \) m\(^{-3}\))
• \( A \) is the cross-sectional area of the wire (\( 2 \times 10^{-6} \) m\(^2\))
• \( e \) is the charge of an electron (\( 1.6 \times 10^{-19} \) C)

We rearrange the formula to solve for \( v_d \):

\( v_d = \frac{I}{nAe} \)

Substitute the provided values:

\( v_d = \frac{3.2}{(8 \times 10^{28})(2 \times 10^{-6})(1.6 \times 10^{-19})} \)

Simplify the calculation:

\( v_d = \frac{3.2}{(25.6 \times 10^3) \times 10^{-13}} \)

\( v_d = \frac{3.2}{25.6 \times 10^{-10}} \)

\( v_d = \frac{3.2 \times 10^{10}}{25.6} \)

\( v_d = 0.125 \times 10^8 = 1.25 \times 10^7 \, \text{m/s} \)

Considering the order of drift speed, we express it as:

\( v_d = 125 \times 10^{-6} \, \text{m/s} \)

The calculated drift speed falls within the given range of 125,125, confirming the solution. Thus, the drift speed of the electrons is \( 125 \times 10^{-6} \, \text{m/s} \).