Question

A 1 m long copper wire carries a current of 1 A. If the cross section of the wire is 2.0 mm² and the resistivity of copper is 1.7 × 10^–8 Ω, the force experienced by moving electron in the wire is ___ × 10^–23N.
(Charge on electron = 1.6 × 10^–19C)

Answer 1

Correct Answer: 136

To determine the force experienced by a moving electron in a copper wire, we use the formula for current density \( J \) and electric field \( E \) in a conductor. The force \( F \) due to an electric field on a charge is \( F = eE \), where \( e = 1.6 \times 10^{-19} \) C is the charge of an electron.
Step 1: Calculate the Current Density 
The current density \( J \) is defined as \( J = \frac{I}{A} \), where \( I = 1 \) A is the current and \( A = 2.0 \) mm² is the cross-sectional area of the wire. Convert the area from mm² to m²:
\( A = 2.0 \, \text{mm}^2 = 2.0 \times 10^{-6} \, \text{m}^2 \).
Thus, \( J = \frac{1}{2.0 \times 10^{-6}} = 5.0 \times 10^5 \, \text{A/m}^2 \).
Step 2: Calculate the Electric Field
Ohm's Law for conductors in terms of current density is \( J = \sigma E \), where \( \sigma = \frac{1}{\rho} \) is the conductivity and \( \rho = 1.7 \times 10^{-8} \, \Omega \, \text{m} \) is the resistivity. Thus,
\( E = J \rho = 5.0 \times 10^5 \times 1.7 \times 10^{-8} = 8.5 \times 10^{-3} \, \text{V/m} \).
Step 3: Calculate the Force on an Electron
The force \( F = eE = 1.6 \times 10^{-19} \times 8.5 \times 10^{-3} = 1.36 \times 10^{-21} \, \text{N} \).
Express the force in the desired format \( F = 1.36 \times 10^{-23} \times 10^2 = 136 \times 10^{-23} \, \text{N} \).
The computed value, \( 136 \), falls within the given range of 136,136. Thus, the force experienced by the electron is within the expected range.

Answer 2

The correct answer is 136
I = nev_dA
\(J=\frac{E}{ρ}\)
\(F=eE=\frac{1.7×1.6×10^{−19}×10^{−8}}{2×10^{−6}}\)
\(=136×10^{–23}N\)