Consider a circuit with some resistance $R$ attached somewhere across a battery. The connection is made using some ideal conducting wires with $0$ resistance. Now, say some current $I$ flows through the wires. This implies there is some current density $\bf J$ flowing through the wires. However, we would require an electric field of $0$ magnitude to achieve this as $\sigma$ is infinite. If that were the case, how exactly is the current flowing? Also, if there was an electric field inside the conducting wire, assuming no resistance, how exactly is the velocity of the electrons (on average) maintained to be a constant? As if they were not constant, the value of the current flowing would change.

Also, what would happen if there was no resistance connected at all? If I suppose the battery doesn't get shorted, hypothetically, how much current should flow through the wires assuming they have $0$ resistance?


2 Answers 2

  1. In practice, $\sigma$ is finite and so we DO require a an electric field to keep the charges moving. This also means that conducting wires are NOT equipotential (think of potentiometers, meter bridges etc.)

  2. As the resistivity of most metals are quite low, we simply NEGLECT the field required to keep them moving (it's there - but we call it zero to simplify calculations). As $\textbf {E} = -\nabla V$, they can also be called equipotential (they're NOT).

  3. The acceleration (and hence velocity) of the electrons is determined by the field applied (the battery) AND the resistance due to the wire. It has been found (experimentally) that for metals and small fields,

$$\textbf {J} = \sigma \textbf{E}$$

which is equivalent to Ohm's law for steady currents.

  1. The electric field (due to the charges or the external source) will NEVER have a component perpendicular to the conductor, as this would imply that charge is leaking out on to the surface(because the field has to be continuous...? - also the external medium is considered to be insulating anyway) -so I'm not really sure about your explanation.

  2. $\sigma \rightarrow \infty$ is just a theoretical limit, so you're bound to run into problems if you try to explain it using physical arguments. Just remember the fact that for real conductors, the applied field will be in the direction of $\textbf{J}$ with NO component along the normals of the conductor. It is just that we let $\textbf{E} = \bf{0}$ as $\sigma \rightarrow \infty$ for perfect conductors.

  • $\begingroup$ This is a great answer. +1 here. $\endgroup$
    – JohnA.
    Commented Jan 4 at 16:37

I feel like I have figured out a possible answer. The electric field is perpendicular to the conducting wires in steady state but until then the conductors are not equipotential and the electrons do face an acceleration through the wires due to the electric field. However, when the surface charges attain a stable configuration, the field inside the conductor becomes $0$ and a constant current is achieved. It is to be noted that Ohm's Law is only valid in steady state and transient state analysis of circuits is not possible by using it.

So in brief, in steady state field lines are perpendicular to the conductor making it equipotential. For real conductors with some resistance, the field inside is never completely $0$ as some amount of resistive force exists which must be overcome by the electric forces.


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