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

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(B) Current density $J$ is defined as $J = I / A$,where $I$ is the current and $A$ is the cross-sectional area.Since the wires are connected in series

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$9 : 1$

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(B) Current density $J$ is defined as $J = I / A$,where $I$ is the current and $A$ is the cross-sectional area.Since the wires are connected in series,the current $I$ flowing through both wires is the same.The cross-sectional area $A$ is given by $A = \pi r^2$.For the copper wire $(1)$: $J_1 = I / (\pi r_1^2)$.For the iron wire $(2)$: $J_2 = I / (\pi r_2^2)$.The ratio of current densities is $J_1 / J_2 = (I / \pi r_1^2) / (I / \pi r_2^2) = r_2^2 / r_1^2$.Given $r_1 = 1 \ mm$ and $r_2 = 3 \ mm$,we have:$J_1 / J_2 = (3)^2 / (1)^2 = 9 / 1$.Therefore,the ratio is $9 : 1$.

Rod	Length,Diameter,Potential Difference
$(A)$	$L, 3d, V$
$(B)$	$2L, d, 2V$
$(C)$	$3L, 2d, 2V$

$A$ copper wire of length $1 \ m$ and radius $1 \ mm$ is connected in series with an iron wire of length $2 \ m$ and radius $3 \ mm$. If a current flows through both wires,the ratio of the current density in the copper wire to that in the iron wire will be:

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