electromagnetism - Does a moving charge interact with the magnetic field it produces?

So the title cuts my question a bit short.

If a moving point charge produces a magnetic field, does the magnetic field it produces affect it?

And let's suppose a moving charge is moving in a region with an externally provided magnetic field, how does the magnetic field produced by the charge's movement interact with/affect/cancel the other magnetic field?

Answer

I think the existing answers are only half right. We need to bring in the fact that the answer depends on the degree to which the charge can be regarded as small, in either amount of charge, or radius, or both.

First let's consider a charge $q$ moving at constant velocity. It is the source of a magnetic field ${\bf B}_q$ in loops around the line of motion. This field has strictly zero net force on the charge that is its source. So in this sense, the answer is "no, the charge does not interact with its own field"---but this is a special case (see rest of this answer). If there is also a further magnetic field ${\bf B}_{\rm ext}$ produced by other currents, then the total field at some place is the vector sum ${\bf B}_q + {\bf B}_{\rm ext}$, but you can't apply this formula right at the location of the charge $q$. The charge $q$ in this case experiences a force $q {\bf v} \times {\bf B}_{\rm ext}$ where $\bf v$ is its velocity.

If the field ${\bf B}_q$ is large enough then it will disturb the motion of other charges and the net result can be that ${\bf B}_{\rm ext}$ is also changed owing to this interaction. However it is common to choose for discussion a 'test charge'. This is one whose charge is small enough that it will not significantly disturb, via its own fields, the motion of anything else.

Now let's come to the case of an accelerating charge. Things get considerably more complicated. Now we have to take into consideration the physical structure of the charged body. It cannot be strictly point-like in classical electromagnetism, because that would lead to infinite fields and infinite mass-energy associated with those fields. In consequence the field due to one part of the charged body can interact with another part of the charged body, and the integral of the resulting force over the whole body (called the self-force) need not be zero. There are now two regimes to think about. If the acceleration is small enough then the self-force is negligible and you can forget about it. This is almost always true in practice, even for particle accelerators. It is only in some extremes of plasma physics and laser physics, or some kinds of particle collision, that this issue is important. Therefore unless one is in such a regime, the answer to the question is still "no" in that we can ignore this slight interaction between the charge and its own field.

However, if the acceleration is large enough that the velocity changes significantly during the time $r/c$ where $r$ is the radius of the body, then the self-force will be non-negligible. It is hard to calculate it exactly, but a good first order approximation for speeds small compared to $c$ is $$ {\bf f}_{\rm self} = \tau_q \frac{d {\bf f}}{d t} $$ where $\tau_q = 2q^2 / 3m c^3$ and $\bf f$ is the force owing to all the other contributions from applied fields. The self-force is often called 'radiation reaction', but strictly that is a slight abuse of terminology in that one can identify a contribution to the self-force that is suitably called radiation reaction, but this is not necessarily the only contribution.

At speeds of any size, the above formula is easily generalized, but this is still a first-order approximation. The equation of motion is $$ m \dot{v}^\mu = f^\mu + \tau_q \left[ \dot{f}^\mu - (\dot{v}_\nu f^\nu) v^\mu/c^2 \right] $$ where $f^\mu$ is the applied four-force and the dot signifies $d/d\tau$ (differentiation with respect to proper time along the worldline). For more information, here is a reference to a couple of papers by myself at Am. J. Phys.: http://dx.doi.org/10.1119/1.4914421; http://dx.doi.org/10.1119/1.4897951 (I mention them since they bear directly on the question asked; I hope that is correct practice).