Question

A copper wire is given having a length $1m$ and radius $1mm$ is connected in series with another wire of iron of length $2m$ and radius $3mm$ . It has been given that the steady current has been passed through this combination. Calculate the ratio of current densities in the copper and iron wires?
$\begin{aligned} & A.18:1 \\\ & B.9:1 \\\ & C.6:1 \\\ & D.2:3 \\\ \end{aligned}$

Answer 1

The current density has been found by taking the ratio of the current passing to the area of the conductor. The area of the cross section can be found by taking the product of the value of pi and the square of the radius. This will help you in answering this question.

Complete step-by-step answer:
The current density has been found by taking the ratio of the current passing to the area of the conductor. This can be written as an equation as,
$J=\dfrac{I}{A}$
The radius of the copper wire has been mentioned as,
${{A}_{1}}=1mm$
The length of the copper wire can be written as,
${{l}_{1}}=1m$
The radius of wire of iron has been mentioned as,
${{r}_{2}}=3mm$
The length of the wire of iron can be written as,
${{l}_{2}}=2m$
The current density of the copper wire can be written as,
${{J}_{1}}=\dfrac{I}{{{A}_{1}}}$
The area of the cross section can be found by taking the product of the value of pi and the square of the radius.
The value of the area of cross section has been written as,
${{A}_{1}}=\pi {{r}_{1}}^{2}$
Substituting the area in the equation of current density as,
${{J}_{1}}=\dfrac{I}{\pi {{r}_{1}}^{2}}$
The current density of the wire of iron can be written as,
${{J}_{2}}=\dfrac{I}{\pi {{r}_{2}}^{2}}$
Taking the ratio of the current density has been given as,
$\dfrac{{{J}_{1}}}{{{J}_{2}}}=\dfrac{{{r}_{1}}^{2}}{{{r}_{2}}^{2}}$
Substituting the values in it will give,
$\dfrac{{{J}_{1}}}{{{J}_{2}}}=\dfrac{{{3}^{2}}}{{{1}^{2}}}=9:1$

So, the correct answer is “Option B”.

Note: The amount of electric current traveling per area of unit cross-section is known as current density. This can be described in amperes per square meter. Higher the current in a conductor, more will be the current density. That is the current density changes in various parts of an electrical conductor.