Question

Two copper wires a and b having lengths in the ratio of 2:1 ,area of cross section in the ratio of 1:2 are connected to a battery of potential difference v separately. What is the ratio of current density a to b

Answer 1

Answer: $\dfrac{J_{a}}{J_{b}} = \dfrac{1}{2}$

Explanation:

Given that,

Ratio of the Length of the copper wires

$\dfrac{l_{a}}{l_{b}} = \dfrac{2}{1}$

Ratio of the area of cross section of the wires

$\dfrac{A_{a}}{A_{b}} = \dfrac{1}{2}$

Using ohm's law

$V = I R$

The current is

$I = \dfrac{V}{R}$....(I)

We know that,

The resistivity is

$R = \dfrac{\rho L}{A}$

Where, $\rho$ is the constant.

Now, put the value of R in equation (I)

Ratio of the current is

$\dfrac{I_{a}}{I_{b}} = \dfrac{VA_{a}}{\rho l_{a}}\times \dfrac{\rho l_{b}}{VA_{b}}$

$\dfrac{I_{a}}{I_{b}} = \dfrac{A_{a}l_{b}}{A_{b}l_{a}}$...(II)

Now, the ratio of current density is

$\dfrac{J_{a}}{J_{b}} = \dfrac{I_{a}\pi r_{b}^{2}}{I_{b}\pi r_{a}^{2}}$

Put the value of I in equation (II)

$\dfrac{J_{a}}{J_{b}} = \dfrac{A_{a}l_{b}\pi r_{b}^{2}}{A_{b}l_{a}\pi r_{a}^{2}}$

$\dfrac{J_{a}}{J_{b}} = \dfrac{1}{2}$

Hence, this is the required solution.