Question

A copper wire of cross-sectional area \[2.0\,{\text{m}}{{\text{m}}^2}\] , resistivity equals \[1.7 \times {10^{ - 8}}\,\Omega \,{\text{m}}\] , carries a current of \[1\,{\text{A}}\] . The electric field in the copper wire is:
A. \[8.5 \times {10^{ - 5}}\,{\text{V}}\,{{\text{m}}^{ - 1}}\]
B. \[8.5 \times {10^{ - 4}}\,{\text{V}}\,{{\text{m}}^{ - 1}}\]
C. \[8.5 \times {10^{ - 3}}\,{\text{V}}\,{{\text{m}}^{ - 1}}\]
D. \[8.5 \times {10^{ - 2}}\,{\text{V}}\,{{\text{m}}^{ - 1}}\]

Answer 1

Hint:First of all, we will find the expression of the resistance and substitute this expression in the Ohm’s law. We will replace potential differences in terms of electric field using a correct relation. After that we will substitute the required values and manipulate accordingly to obtain the result.

Complete step by step answer:
In the given question, we are supplied the following data:
There is copper wire whose area of cross section is \[2.0\,{\text{m}}{{\text{m}}^2}\]. The resistivity of the copper wire is \[1.7 \times {10^{ - 8}}\,\Omega \,{\text{m}}\]. The current flowing through the copper wire is \[1\,{\text{A}}\]. We are asked to find the electric field in the copper.

To begin with, we will analyse the physical quantities that are given and the physical quantities that are required to be found. We see that the resistance is not given in the question and we are required to find it first, in order to proceed to find the electric field.Let us proceed to find the resistance of the wire. The resistance of the wire is given by the expression which is given below:
\[R = \rho \times \dfrac{l}{A}\] …… (1)
Where,
\[R\] indicates the resistance of the wire.
\[\rho \] indicates the resistivity of the wire.
\[l\] indicates the length of the wire.
\[A\] indicates the area of the cross section of the wire.

We know Ohm’s law which is given by:
\[V = i \times R\]
\[ \Rightarrow V = i \times \rho \times \dfrac{l}{A}\] …… (2)

Again, we know that the electric field is given by:
\[V = E \times l\] …… (3)
Where,
\[V\] indicates potential difference.
\[E\] indicates an electric field.

Now, we substitute the equation (3) in equation (2) and we get:
$E \times l = i \times \rho \times \dfrac{l}{A} \\
\Rightarrow E = i \times \rho \times \dfrac{1}{A} \\
\Rightarrow E = 1 \times 1.7 \times {10^{ - 8}} \times \dfrac{1}{{2 \times {{\left( {1 \times {{10}^{ - 3}}} \right)}^2}}} \\
\therefore E = 8.5 \times {10^{ - 3}}\,{\text{V}}\,{{\text{m}}^{ - 1}} \\
\\$
Hence, the electric field in the copper is \[8.5 \times {10^{ - 3}}\,{\text{V}}\,{{\text{m}}^{ - 1}}\] .The correct option is C.

Note:While solving this problem, most of the students forget to convert the unit of the area of the cross section, which is square millimetres to square metres. Students get wrong and irrelevant answers, due to this reason. The electric field in the copper increases when there is an increase in the current through the wire.