References: Griffiths, David J. (2007), Introduction to Electrodynamics, 3rd Edition; Pearson Education – Problem 10.19.
In his example 10.4, Griffiths works out the fields due to a point charge moving with constant velocity . They are
is the vector from the particle’s present (not retarded) position to the observer (assuming the particle passes through the origin at ) and is the angle between and . We can use this formula to rederive the equation for the electric field due to an infinite line charge with linear charge density . From electrostatics, we know the field is given by
where is the perpendicular distance from the line (wire). Let’s see if we can get the same result using the formula above.
The field due to a small segment of the wire of length at position is that due to a point charge . For an observation point at , the length of is
and since the velocity is parallel to the wire, we have
Since is parallel to , by symmetry the components of parallel to the wire will cancel out, since there will be equal and opposite contributions from points . The perpendicular component is so the total field is
where the direction is radial. We can convert this to an integral over by noting that
The integral can be evaluated using Maple, and we get
so we get back the correct field
The magnetic field of a point charge is given by Griffiths as
Since is a constant, the total magnetic field can be found from the same integral as above. Its direction is given by which circles the wire in a direction given by the usual right-hand rule. Since (the current), we get
which agrees with the magnetostatic formula using Ampère’s law.