Question

A copper wire of length 1m and radius $1\text{mm}$ is joined in series with an iron wire of length $2\text{m}$ and radius $3\text{mm}$, and a current is passed through the wires. The ratio of the current density in the copper and iron wires is:

• A. $18:1$
• B. $9:1$
• C. $6:1$
• D. $2:3$

Answer 1

The correct option is B $9:1$

Given:

$l_1=1\text{m},r_1=1\text{mm}$

And $l_2=2\text{m},r_2=3\text{mm}$

As we know that the current density is given by,

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

and $I$ is same in series,

$A=\pi r^2$

Therefore, the ratio of current density is,

$\frac{J_1}{J_2}=\frac{\frac{I}{A_1}}{\frac{I}{A_2}}=\frac{A_2}{A_1}$

$\frac{J_1}{J_2}=\frac{\pi (3{)}^2}{\pi (1{)}^2}=9$