A copper wire of length 1, m and radius 1, mm | vedclass

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(B) Current density $J$ is defined as the current per unit cross-sectional area,given by $J = \frac{I}{A} = \frac{I}{\pi r^2}$.Since the copper wire a

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$9:1$

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(B) Current density $J$ is defined as the current per unit cross-sectional area,given by $J = \frac{I}{A} = \frac{I}{\pi r^2}$.Since the copper wire and the iron wire are connected in series,the same amount of current $I$ flows through both wires,i.e.,$I_{Cu} = I_{Fe} = I$.The ratio of the current density in the copper wire $(J_{Cu})$ to the iron wire $(J_{Fe})$ is given by:$\frac{J_{Cu}}{J_{Fe}} = \frac{I / (\pi r_{Cu}^2)}{I / (\pi r_{Fe}^2)} = \frac{r_{Fe}^2}{r_{Cu}^2}$.Given $r_{Cu} = 1\, mm$ and $r_{Fe} = 3\, mm$,we substitute these values:$\frac{J_{Cu}}{J_{Fe}} = \frac{(3\, mm)^2}{(1\, mm)^2} = \frac{9}{1} = 9:1$.Thus,the ratio of the current density is $9:1$.

$A$ 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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