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} $