Question

A copper wire of length $1m$ and radius $1mm$ is connected in series with another wire of iron of length $2m$ and radius $3mm$. A steady current is passed through this combination. The ratio of current densities in copper and iron wires will be
(A) $18:1$
(B) $9:1$
(C) \[6:1\]
(D) \[2:3\]

Answer 1

Hint: To solve this question, we need to use the formula for the current density. The current density is equal to the ratio of the current and the cross sectional area of the conductor through which the current is flowing normally. We have to find out the current densities for the two wires, keeping the current same for their serial combination.
Formula used: The formula used to solve this question is given by
$J = \dfrac{I}{A}$, here $J$ is the current density, $I$ is the current flowing through a conductor having an area of cross section of $A$.
$A = \pi {r^2}$, here $A$ is the area of the cross section of a wire having a radius of $r$.

Complete step-by-step solution:
Let the steady current passed through the combination of the copper and the iron wire be equal to $i{\text{ A}}$.
We know that the current through the elements which are connected in a serial combination is the same. So the current through each of the copper and the iron wires will be equal to $i{\text{ A}}$.
Now, we know that the current density is defined as the current flowing through a conductor per unit cross sectional area. So it is given by
$J = \dfrac{I}{A}$ ….(1)
Also, we know that the area of cross section of a wire is given by
$A = \pi {r^2}$....(2)
Putting (2) in (1) we get
$J = \dfrac{I}{{\pi {r^2}}}$...................(3)
According to the question, the radius of the copper wire is equal to $1mm$. Therefore, on substituting $r = 1mm$ and \[I = i{\text{ A}}\] in (3) we get the current density in the copper wire as
\[{J_C} = \dfrac{i}{{\pi {{\left( 1 \right)}^2}}}{\text{A}}m{m^{ - 2}}\]
$ \Rightarrow {J_C} = \dfrac{I}{\pi }$ …………..(4)
Also, the radius of the iron wire is given equal to $3mm$. Therefore, substituting $r = 3mm$ and $I = i$ in (3) we get the current density in the iron wire as
${J_I} = \dfrac{i}{{\pi {{\left( 3 \right)}^2}}}{\text{A}}m{m^{ - 2}}$
$ \Rightarrow {J_I} = \dfrac{i}{{9\pi }}{\text{A}}m{m^{ - 2}}$............(5)
Dividing (4) by (5) we get
\[\dfrac{{{J_C}}}{{{J_I}}} = \dfrac{{\dfrac{i}{\pi }}}{{\dfrac{i}{{9\pi }}}}\]
\[ \Rightarrow \dfrac{{{J_C}}}{{{J_I}}} = \dfrac{{9\pi }}{\pi } = 9\]
Thus, the required ratio is equal to $9:1$.

Hence, the correct answer is option B.

Note: The values given for the length and the area of the cross section of the wires may tempt us to use the formula for the resistance which is given by $R = p\dfrac{l}{A}$. But since we have no idea regarding the resistivity of the wires, this formula cannot be used here