# Charge moving with parallel to current with same velocity as particles in the wire

I have a charge $+Q$ traveling paralel to a wire, with velocity $\vec v$. The wire is composed of negative non moving charge and a flow of positive charges moving with velocity, $\vec v$, with a spacing $l$ between them.

I am looking at the one on the diagram on the left right now.

The electric field is $$\vec E(r) = \frac{q}{2\pi l r \epsilon_0}\hat r.$$

The magnetic force is $$\oint \vec B \vec dl = \mu_0 I_\text{encl} = \mu_0 \lambda\vec v=\mu_0 \vec v \frac{q}{l}\\ \Rightarrow \vec B = \frac{\mu_0q}{2\pi l}\vec v.$$ The lorentz force is $$\vec F = Q\left[ \vec E + (\vec v \times \vec B)\right]$$

Because the magnetic fieldis in the same direction at the moving changed particle is $$(\vec v \times \vec B) = 0 \\ \Rightarrow \vec F = Q\vec E?$$

• Interesting question. – Declan Jun 5 '18 at 4:39
• Sorry. I had more to add. Can you now calculate the particle trajectory in the lab frame? – Declan Jun 5 '18 at 4:40

On the left the charge is moving, the wire is neutral with a current: $\vec F = Q \vec v \times \vec B = \frac{\mu_0 Q v I}{2\pi r}$.
On the right the charge is at rest, so not affected by $\vec B$. However, the wire carries a charge density $\lambda = \vec v \cdot \vec I /c^2$ in this reference frame. The force is the same as before $\frac{Q\lambda}{2\pi c^2 \epsilon_0 r } = \frac{QvI}{2\pi c^2 \epsilon_0 r} = \frac{\mu_0 Q v I}{4\pi r}$ .