# Does the direction of electric and magnetic field change in a circuit?

From my understanding, when studying a circuit, the magnetic field is either "in" the page or "out" the page according to whether the current flows clockwise or anticlockwise. So that direction should be fixed. Same goes for the electric field, which is either top-bottom or bottom-top, parallel to the sheet, from the place where the positive potential is to where the negative is.

This seems straightforward enough. My question is do their directions ever change at different points in the circuits? I would argue the magnetic field does not change because the direction of the current is the same. But what about the electric field? I am really not sure, as the positive and negative potential remains in the same position, but maybe the current as transport of charges influences its direction locally?

Your intuition for the direction of the electric and magnetic fields around a circuit should always come from the position and movement of charges.

If you look at a battery by itself, there are electric field lines that travel from its anode to its cathode. Charge carriers want to follow these field lines, but can't because the resistance of air is too high. (That is, because electrons can't easily rearrange themselves among the atoms in air, in order to move a (negative) charge from the cathode to the anode, you would have to dissociate an electron from an atom somewhere, forming an positive ion and electron. However, that highly positive ion will now strongly attract the electron back, so self-ionization like this only happens with very strong applied electric field.)

When you connect a battery to a conductor that closes the loop in the circuit, charges can now travel easily through the conductor, and follow the electric field lines to do so. However, the act of these charges moving actually changes the field!

What happens is that charges rapidly build up on the surface of the conductor to largely cancel the electric field inside the conductor, but a small field is left because the conductor has finite conductance. Adding resistors to the circuit also causes charges to pile up on both sides of the resistor, causing increased electric field strengths through and thus voltage drop across the resistor. Throughout the circuit, electrons feel whatever residual field is left, and drift to follow it, moving on average from the cathode to the anode, producing a net current.

If you want to look at the electric and magnet fields outside the circuit, there are two things to look at:

1. The surface charges on the conductor and lumped elements will source fields. In steady-state, the surface charges don't move, so they only contribute to electrostatic fields.
2. The drifting charges within the conductor and surface elements will produce magnetic fields, because they are moving. In steady-state, the current is constant, so these charges contribute to the magnetostatic field.

Outside of steady-state, you still should consider how the drift current and surface charges are changing in order to understand what the field looks like, but the situation can be more complicated.

An in-depth explanation of this comes from the J. D. Jackson in his 1996 paper "Surface charges on circuit wires and resistors play three roles," but I can share perhaps more introductory resources in the comments if need be.

Your analysis is wrong. For a picture showing an example, see http://lightandmatter.com/fac/ (I'm the author), ch. 8, figure n.

The magnetic field vectors don't all have the same direction. They form closed loops around the wires. The field very close to a wire is what you would expect for the field of a long, straight wire.

If the conductor is a perfect conductor, then the electric field at the surface of the wire is always perpendicular to the wire, which is different from what you say. (My diagram is actually for the more realistic case where the wires have some small resistance, so the electric field is only approximately perpendicular to the surface of the wire.)

The answer by Jonathan Jeffrey says:

When you connect a battery to a conductor that closes the loop in the circuit, charges can now travel easily through the conductor, and follow the electric field lines to do so.

This is wrong. If it's a perfect conductor, then the electric field is zero inside the conductor. In a real-world circuit, the current in a wire normally flows very nearly perpendicular to the electric field. (If this were not the case, then you would get power dissipation in the wire, which does happen but is normally a negligible effect.)

• Perfect conductors don't exist in the real world. My answer described precisely how charges rearrange themselves to travel to the surface of a conductor to cancel almost all of the field in the conductor. However, because the conductor has finite conductance (nonzero resistance), there will be small electric field somewhere in the conductor, and this is precisely the field that drives drift current. Commented Feb 1, 2021 at 15:16
• Now, you might be correct that current flows nearly perpendicularly to the electric field (e.g. if the field is nearly orthogonal to the conductor surface) but my answer made no mention of the actual direction of the small electric field within the wire. It is not necessary to precisely find that to get the intuition. Commented Feb 1, 2021 at 15:19
• But I do agree that you can think of most current of the electrons as being due to momentum from their original acceleration before surface charges settle, but the residual field is required to keep the charges going: after all, room-temperature wires do not form superconducting loops. Commented Feb 1, 2021 at 15:29