# Interaction between a current-carrying wire and charge moving at right angle to the wire

**Interaction between a current-carrying wire and charge moving at right angle to the wire**

Edward M. Purcell Electricity and Magnetism, Berkeley Physics Course, Vol. 2 in SI units

Charge moving perpendicular to the wire experiences a force parallel to the wire, perpendicular to its direction of motion. Positive ions cannot cause a horizontal field at the test charge position. The x' component of the field from
an ion on the left is exactly cancelled by the x' component of the field of a symmetrically positioned ion on the right. The effect we can see is caused by the electrons. Electrons are all moving diagonally in the test charge frame,
downward and toward the right. Consider the two symmetrically located electrons e_{1} and e_{2}. Their electric fields, relativistically compressed in the direction of the electrons' motion, are represented by a field
lines. You can see that, although e_{1} and e_{2} are equally far away from the test charge, the field of electron e_{2} is stronger than the field of electron e_{1} at that location. That is
because the line from e_{2} to the test charge is more nearly perpendicular to the direction of motion of e_{2}. In other words, the angle θ' that appears in the denominator of equation for the magnitude of the
field of electron in the frame moving relative to the electron is here different for e_{1} and e_{2}, so that sin^{2}θ'_{2} > sin^{2}θ'_{1}. That is true for
any symmetrically located pair of electrons on the line, as you can verify with the aid of picture. The electron on the right always acts stronger. Therefore, summing over all the electrons must give a resultant field E' in the
** x** direction. The y' component of the electrons' field is exactly cancelled by the field of the ions. That E'

_{y}is zero is guaranteed by Gauss's law, for the number of charges per unit length of wire is the same as it was in the lab frame. The wire is uncharged in both frames. The force on the test charge, qE'

_{x}, transformed back into the lab frame is a force proportional to v in the

**direction, which is the direction of**

*x***if B is a vector in the**

*v×B***direction, pointing out of the page.**

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