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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(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 $\frac{{{J_1}}}{{{J_2}}} = \frac{{r_2^2}}{{r_1^2}} = 9:1$
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