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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

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18 : 1

9 : 1

6 : 1

2 : 3

answer is B.

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Detailed Solution

Current density $J = \frac{i}{A} = \frac{i}{{πr}^{2}} \Rightarrow \frac{J_{1}}{J_{2}} = \frac{i_{1}}{i_{2}} \times \frac{r_{2}^{2}}{r_{1}^{2}}$

But the wires are in series, so they have the same

current, hence $i_{1} = i_{2} . So \frac{J_{1}}{J_{2}} = \frac{r_{2}^{2}}{r_{1}^{2}} = 9 : 1$

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