Topic: Intro to Current. Two cylindrical copper wires have the same length but different diameters. Wire A has diameter $d_A$ and Wire B has diameter $d_B$, with $d_B>d_A$. If the same potential difference $V$ is applied across each wire, which wire can carry the higher current?

• A. Both carry the same current because the material is copper in both cases.
• B. Wire B, because its larger cross-sectional area gives a smaller resistance $R=\frac{\rho L}{A}$ and thus a larger current $I=\frac{V}{R}$.
• C. Neither; current depends only on length $L$, not on diameter.
• D. Wire A, because a smaller diameter increases the electric field and therefore increases current.

Answer: (B)

Verified step by step guidance

1. Identify the key variables: both wires have the same length \( L \), different diameters \( d_A \) and \( d_B \) with \( d_B > d_A \), and the same potential difference \( V \) applied across them.
2. Recall the formula for the resistance of a cylindrical wire: \( R = \frac{\rho L}{A} \), where \( \rho \) is the resistivity of the material (same for both wires since both are copper), \( L \) is the length, and \( A \) is the cross-sectional area.
3. Express the cross-sectional area \( A \) of each wire in terms of its diameter: \( A = \pi \left( \frac{d}{2} \right)^2 = \frac{\pi d^2}{4} \). Since \( d_B > d_A \), wire B has a larger cross-sectional area.
4. Use Ohm's law to relate current \( I \), voltage \( V \), and resistance \( R \): \( I = \frac{V}{R} \). A smaller resistance means a larger current for the same voltage.
5. Combine these insights: wire B has a larger cross-sectional area, so its resistance \( R_B \) is smaller than \( R_A \), leading to a larger current \( I_B \) compared to \( I_A \) when the same voltage \( V \) is applied.