Question

A copper wire of length 1m and radius 1mm is connected in series with another wire of iron of length 2 m and radius 3 mm. 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

Answer: (B)

Step 1: Given that

Length of copper wire= $1m$

Radius of copper wire= $1mm$ = $\frac{1}{1000}m=0.001m$

The length of iron wire= $2m$

Radius of the iron wire= $3mm=0.003m$

Step 2: Determination of the ratio of the current density of copper wire to the iron wire:

The current density of a conductor is given by,

$j=\frac{I}{A}$

Where$\text{I}$is the current in the conductor and $\text{A}$ is the area of cross-section of the conductor.

As the conductors are connected in series, each will have the same flow of electric current.

Let $I$ be the electric current in both the conductors then,

Current density of copper wire is given by

$j_{copper}=\frac{I}{\pi {\left(r_{copper}\right)}^2}$

$j_{copper}=\frac{I}{\pi {\left(0.001m\right)}^2}$

$j_{copper}=\frac{I}{\pi \times \frac{1}{1000\times 1000}}$

$j_{copper}=\frac{{10}^{6}I}{\pi }$

And current density of iron is given by;

$j_{iron}=\frac{I}{\pi {\left(0.003m\right)}^2}$

$j_{iron}=\frac{I}{\pi \times \frac{9}{1000000}}$

$j_{iron}=\frac{{10}^{6}I}{9\pi }$

Now,

$\frac{j_{copper}}{j_{iron}}=\frac{{10}^{6}I}{\pi }\times \frac{9\pi }{{10}^{6}I}$

$\frac{j_{copper}}{j_{iron}}=\frac{9}{1}$

$j_{copper}:j_{iron}=9:1$

Thus,

Option A) 9:1 is the correct option.