The potential difference between two ends of a 45-cm length of copper wire (1.7 $\times$ 10^-8 $\Omega$·m) is 1.5 V. The magnitude of the current density in the wire is A) 1.0 $\times$ 10⁸ A/m². B) 1.5 $\times$ 10⁸ A/m². C) 2.0 $\times$ 10⁸ A/m². D) 2.5 $\times$ 10⁸ A/m².

Question

Quizplus · Accepted Answer

The current density $ J $ is given by $ J = \frac{V}{\rho \cdot L} $, where $ V = 1.5 \, \text{V} $, $ \rho = 1.7 \times 10^{-8} \, \Omega \cdot \text{m} $, and $ L = 0.45 \, \text{m} $. Calculating gives $ J = \frac{1.5}{1.7 \times 10^{-8} \times 0.45} = 2.0 \times 10^8 \, \text{A/m}^2 $.

The Potential Difference Between Two Ends of a 45-Cm Length ×\times×

The Potential Difference Between Two Ends of a 45-Cm Length $\times$