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In Example 7.9 (Griffiths Introduction to Electrodynamics), I know how to determine direction of current due to induced electric field (see example in the figure below), but I have trouble understanding why the solution says "E here runs parallel to the axis". Why does the induced electric field have no vertical component? Could anyone provide mathematical proof for that?

Example 7.9:
An infinitely long straight wire carries a slowly varying current I(t). Determine the induced electric field, as a function of the distance s from the wire. FIGURE 7.27 Solution
In the quasistatic approximation, the magnetic field is $\frac{\mu_0 I}{2 \pi s}$, and it circles around the wire. Like the B-field of a solenoid, E here runs parallel to the axis. For the rectangular “Amperian loop” in Fig. 7.27, Faraday’s law gives: $$\oint \vec{E} \cdot d\vec{l} = E(s_0)l - E(s)l$$ ... ...

Tan En De
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3 Answers3

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There is a counterclockwise emf induced in the loop but the symmetry of the loop means that any electric field in the "vertical" parts of the line integrals cancel each other out.
To illustrate this suppose the magnitude of the electric field electric field, $E$, was constant on the two vertical sides.
The two parts of the line integral would be $E(s-s_0)$ and $E(s_0-s)$ which when added together give zero.

Farcher
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We can use a symmetry argument. The rule is as follows:

At a point on a plane of symmetry, polar vectors such as the electric field are contained in the plane and axial vectors, such as the magnetic field, are normal to the plane.

For an antisymmetry plane (system invariant after symmetry and a change in the signs of the charges) it is the opposite.

Here, the plane normal to the wire passing through $M$ is a plane of antisymmetry: if we make a plane symmetry and reverse the sign of the charges, we find the same situation. The electric field is therefore normal to the plane.

We can justify it in a little more detail: if we had a radial component of the field, it would be unchanged after plane symmetry and would change sign after the change in sign of the charges. We would therefore have a radial component with a changed sign. But the system is invariant after these two operations and as the solution is unique, it must be zero.

Hope it can help and sorry for my poor english.

Vincent Fraticelli
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The radial electric field is zero because of the assumptions he is making: An infinitely long wire with uniform current varying only in time, and the uniform current is the only source of electric field. There is no charge on the wire, and the situation is symmetric along the wire and around the wire, so there cannot be a radial electric field.

You do not in general need to make those assumptions. For example, suppose the wire is merely very long, not infinite. Think many miles long at the same scale as the OP's image. Lets further assume that the current is due to a freely oscillating current on the wire where the current distribution along the wire has one full period of a sinusoid. The peak current occurs 1/4 of the way in from each end, and the charge density goes through a zero at those points. Griffith's example is an approximation of the situation at those two points.

But you could also ask what is the electric field at the center of the very long wire. There, the current is approximately zero, and the charge is approximately constant along the wire, varying only in time. Then the electric field would be only radial. Looking at the contribution of the radial electric field along the two sides to the EMF around the loop, they cancel out, leaving zero EMF at the wire center. So any electric field due to charge on the wire doesn't affect his final answer anyway.

Finally, if you wanted the electric field at a general location along the wire, you would need both of those contributions, with the radial component coming from the charge at that point on the wire.

BaddDadd
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