The direction of induced electric field around a changing current wire

Question

I saw in my textbook that the induced electric field around a changing current wire is parallel to the direction of the current, for example:

Why is that? Thanks!

Answer 1

This is true only close enough to the wire; when you get farther so that other parts of the circuit are as close as the original element of wire, induced field can point in any direction, because circuit is a closed loop and all parts of the circuit where current changes influence the induced field.

When you have a long straight wire and measure close to it, then indeed induced electric field is in direction of the wire. This is because induced field is due to acceleration of charged particles in the wire, and accelerated charged particle produces radiation field vector that is both 1) perpendicular to the line of sight towards the particle, 2) lies in the plane defined by the line of sight and acceleration.

So when close to the wire, all such accelerated charged particles produce additional electric fields proportional to their acceleration, and with direction defined above. All these fields add up (vectorially) to the only possible direction that is exactly or almost the same as that of the wire ("almost" due to the fact that the straight wire can't be infinitely long, there is eventually a turn and this may cause the total induced field to deviate slightly from direction of the wire).

Why is electric field of an accelerated charge such? This follows from Maxwell's equations, and these in turn are generalization of many experiments with EM phenomena.

The fact the induced field is perpendicular to the line of sight can be nicely illustrated by the Thomson construction of field lines of a charge that undergoes a brief period of acceleration. See the animation here:

http://www.tapir.caltech.edu/~teviet/Waves/empulse.html

As you can see, the acceleration produced kinks in electric field lines; in those kinds, electric field direction is modified and has a component perpendicular to the radial line.

The fact that the field lines are continuous outside the source charge, they do not break or that new lines do not appear in the vacuum, is due to single Maxwell equation

$$ \nabla \cdot \mathbf E = 0, $$

also known as the Gauss law for electric field in vacuum.

This law alone implies that the field due to acceleration will have kinks and thus the field will have perpendicular "radiation"-like component. The fact this component is proportional to acceleration is due to the other Maxwell equations.

Answer 2

An electric field is what drives charge to move, since ${\bf E}$ is a force per unit charge (${\bf F}=q{\bf E}$). Current (motion of charge) will flow in the direction of an electric field. It works the other way as well, if there is a current, there must be an electric field in that direction driving it (for example from an electric potential difference between two ends of the wire, giving rise to an ${\bf E}$ field through ${\bf E}=-\nabla V$).

Often for simple wires, you can make the relation between current density and electric field: ${\bf J}=\sigma{\bf E}$, where $\sigma$ is the conductivity. This is Ohm's law.