Two copper wires A and B, of diameters 1 mm and 3 mm respectively, are joined end to end, and a steady current flows through the combination. The ratio of the electric field inside wire A to that inside wire B is
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → 9 : 1
The two wires are joined end to end, so the same current \( I \) flows through each. Inside a conductor the field and the current density are related by Ohm’s law in the form \( E=\rho j \), with \( j=I/A \). Both wires are of copper, so \( \rho \) is the same and the field depends only on the area of cross-section.
For each wire:
\( \displaystyle E=\frac{\rho I}{A} \)
With \( A=\pi d^{2}/4 \), the field varies as \( 1/d^{2} \). The ratio of the fields is:
\( \begin{aligned} \frac{E_A}{E_B} &= \frac{d_B^{2}}{d_A^{2}} \\ &= \frac{(3\,\mathrm{mm})^{2}}{(1\,\mathrm{mm})^{2}} \\ &= 9 \end{aligned} \)
Therefore, \( E_A:E_B=9:1 \). The field is stronger in the thinner wire, because the same current flows through a smaller area.