Question

A copper wire of length $1\, m$ and radius $1\, mm$ is joined in series with an iron wire of length $2\, m$ and radius $3\, mm$ and a current is passed through the wires. The ratio of the current density in the copper and iron wires is

• A. 18 : 1
• B. 9 : 1
• C. 6 : 1
• D. 2 : 3

Answer 1

Answer: (B)

Current density $J=\frac{i}{A}=\frac{i}{\pi r^{2}}$
$\Rightarrow \frac{J_{1}}{J_{2}}=\frac{i_{1}}{i_{2}} \times \frac{r_{2}^{2}}{r_{1}^{2}}$
But the wires are in series, so they have the same current, hence, $i_{1}=i_{2}$
So, $\frac{J_{1}}{J_{2}}=\frac{r_{2}^{2}}{r_{1}^{2}}$
$=9: 1$