Relation between Current and Drift Velocity: Practice Problem & Solution
A copper wire of length $10\text{ m}$ and radius $(10^{-2}/\sqrt{\pi})\text{ m}$ has electrical resistance of $10\ \Omega$. The current density in the wire for an electric field strength of $10\text{ V/m}$ is: (2022)
Solution Explained:
To solve this problem, we apply the core principles of Relation between Current and Drift Velocity. Understanding the underlying formula is key to arriving at the correct answer below:
Area $A = \pi r^2 = \pi \left(\frac{10^{-2}}{\sqrt{\pi}}\right)^2 = 10^{-4}\text{ m}^2$.
Resistivity $\rho = \frac{R A}{L} = \frac{10 \times 10^{-4}}{10} = 10^{-4}\ \Omega \cdot \text{m}$.
Current density $J = \frac{E}{\rho} = \frac{10}{10^{-4}} = 10^5\text{ A/m}^2$.
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