Question

A copper wire of resistivity $1.63\times {{10}^{-8}}\Omega m$ is having a cross section area mentioned to be $10.3\times {{10}^{-4}}c{{m}^{2}}$. What will be the length of the wire required in order to make a $20\Omega$ coil?

Answer 1

The resistance of the copper wire can be found by taking the ratio of the product of the resistivity of the copper wire and the length of the copper wire to the area of the cross section of the copper wire. Substitute the values in the equation. Then rearrange it in terms of the length of the wire. This will help you in answering this question.

Complete answer:
It has been given that the resistivity of the copper wire is,
$\rho =1.63\times {{10}^{-8}}\Omega$
The area of the cross section of the copper wire has been mentioned as,
$A=10.3\times {{10}^{-4}}c{{m}^{2}}$
The resistance connected can be written as,
$R=20\Omega$
The resistance of the copper wire can be found by taking the ratio of the product of the resistivity of the copper wire and the length of the copper wire to the area of the cross section of the copper wire. This can be written as an equation given as,
$R=\dfrac{\rho l}{A}$
Substituting the values in this equation will give the resistance of the copper wire. That is we can write that,
$20=\dfrac{1.63\times {{10}^{-8}}l}{10.3\times {{10}^{-4}}\times {{10}^{-4}}}$
Rearranging this equation can be shown as,
$l=\dfrac{20\times 10.3\times {{10}^{-8}}}{1.63\times {{10}^{-8}}}=126.4m$
Therefore the length of the copper wire has been calculated. The correct answer has been found out.

Note:
Resistivity can be defined as the electrical resistance of a conductor or a metal wire having a unit cross-sectional area and a unit length. This is a characteristic feature of each substance. The resistivity will be helpful in the comparison of different materials on the basis of their quality to conduct electric currents. High resistivity will show that the material is a bad conductor.