# Effect of a current-carrying infinite wire moving towards a stationary charge

An infinite wire carrying electrical current moving towards a stationary charge (perpendicular direction), why is there an electrical force on the charge in a direction parallel to the wire? I can only work out a force perpendicular to the wire, no parallel component. (Note in the wireframe, charge moves perpendicular towards the wire, so it must feel a magnetic force parallel to the wire)

In a problem like this which is interested in electromagnetism in different frames, the best approach is to use four-vectors and tensors. The charge has a four-current (in natural units) $$q^{\mu }=\left(q,0,0,0\right)$$ in the unprimed frame.

In the primed frame, place the wire on the $$z'$$ axis with the current in the positive direction. Then the electromagnetic field tensor is given by $$F^{\mu \nu }=\left( \begin{array}{cccc} 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & \frac{j x'}{x'^2+y'^2} \\ 0 & 0 & 0 & \frac{j y'}{x'^2+y'^2} \\ 0 & -\frac{j x'}{x'^2+y'^2} & -\frac{j y'}{x'^2+y'^2} & 0 \\ \end{array} \right)$$

When we transform the electromagnetic field to the unprimed frame we get $$F^{\mu \nu }=\left( \begin{array}{cccc} 0 & 0 & 0 & -\frac{j v (t v+x)}{(t v+x)^2-\left(v^2-1\right) y^2} \\ 0 & 0 & 0 & \frac{j (t v+x)}{(t v+x)^2-\left(v^2-1\right) y^2} \\ 0 & 0 & 0 & \frac{j y}{y^2-\frac{(t v+x)^2}{v^2-1}} \\ \frac{j v (t v+x)}{(t v+x)^2-\left(v^2-1\right) y^2} & -\frac{j (t v+x)}{(t v+x)^2-\left(v^2-1\right) y^2} & -\frac{j y}{y^2-\frac{(t v+x)^2}{v^2-1}} & 0 \\ \end{array} \right)$$

Notice that in the unprimed frame there is an E field, not just a B field. This can be understood by the fact that in the primed frame the B field is changing as the wire moved. This changing B field induces an E field.

The four-force on the charge is given by $$f_{\mu }=F_{\mu \nu } q^{\nu }=\left(0,0,0,-\frac{j q v (t v+x)}{(t v+x)^2+\left(1-v^2\right) y^2}\right)$$ where you can see that there is a force in the $$z$$ direction, which is the direction parallel to the wire.

Because, as you've pointed out, in the frame of the wire, the charge experiences non-zero magnetic force in direction of the wire. Since in this frame there is no electric force, due to how EM fields transform, in the frame of the charge, there has to be a non-zero force in direction of the wire too. This force can only be due to electric field, because magnetic force is zero, because the charge is at rest.

The electric force is due to electric field of the moving wire. This electric field can be determined from the magnetic field in the frame of the wire using the Lorentz transformation. In general, any moving magnet, electromagnet or even a piece of current-carrying wire carries around itself a non-zero electric field.

There is no electrical force on the charge as the wire is electrically neutral.

If the charge is moving perpendicularly towards the wire, there will be a magnetic force. You can use fleming's left hand rule to figure out the direction of the force.

The Force vector is given by:
F = q(v x B)
(v is velocity of charge, B is magnetic field at the position of charge. The cross product of v and B gives direction of Force)

The direction of force is parallel or antiparallel depending whether the charge is positive or negative.