Question

A copper wire of length $L$ and cross sectional area $A$ carries a current $I$ . If the specific resistance of copper is $S$ , then what is the electric field in the wire.
(A) $ISA$
(B) $\dfrac{{IA}}{S}$
(C) $\dfrac{{IS}}{A}$
(D) $\dfrac{A}{{SI}}$

Answer 1

Hint : The electric field in the wire can be expressed in voltage per unit length. The specific resistance is the same as the resistivity of a conductor. So from the given values in the question we can derive the formula for the electric field as voltage per unit length.

Formula used: In this solution we will be using the following formula;
$E = \dfrac{V}{L}$ where $E$ is the electric field, $V$ is the potential difference or voltage, and $L$ is length.
$S = \dfrac{{RA}}{L}$ where $S$ is the specific resistance, $R$ is resistance, $A$ is cross section area and $L$ again is length.
$V = IR$ where $I$ is the current in the conductor.

Complete step by step answer
To get the electric field in the wire, we shall note that a copper wire obeys ohm’s law. Hence,
$V = IR$ where $I$ is the current in the conductor, $V$ is the potential difference or voltage and $R$ is resistance.
Now resistance can be given by
$R = \dfrac{{SL}}{A}$ where $S$ is the specific resistance, $A$ is cross section area and $L$ is length.
Thus, by inserting the expression into above equation, we get
$V = I\dfrac{{SL}}{A}$
From prior knowledge, we know that the voltage can be given by
$V = EL$ , hence by replacing into the above equation, we have that
$EL = I\dfrac{{SL}}{A}$ , cancelling the length from both sides, we have that
$E = \dfrac{{IS}}{A}$
Hence, the correct answer is option C.

Note
Alternatively, a simple dimensional analysis can give our answer. We perform a unit check as follows
We know that the unit of electric field can be given as volts per meter, from $E = \dfrac{V}{L}$ .
But the Volts from $V = IR$ can also be given as ampere-Ohms, hence the unit of electric field can be given as Ampere-ohms per meter.
Now since the unit of $S$ is Ohms-meter
Then the right arrangement for electric field is
$E = A\dfrac{{\Omega m}}{{{m^2}}} = A\Omega /m$
Thus,
$E = \dfrac{{IS}}{A}$