I Does electron beam in empty space generate magnetic fields?

[avicenna]

Does electron beam in empty space generate magnetic fields around them just as with current in conductor.

If yes, then is it experimentally proven that two parallel electron beam would attract each other.

 

[BvU]

Yes to the first one.
How do you come to expect attraction in your second question?

$\$

 

[Delta2]

Yes to the first one from me too. But for the second one , it isn't so simple if they attract or repel each other because yes there is one component of the total force , the magnetic force (call it laplace or lorentz force) that is attractive, but there is also the electrostatic coulomb force between the electrons streams that is repulsive.

Between conductors the repulsive force between the electrons streams is neutralized by the attractive force of the positive ions, or in simple word the conductors are electrically neutral, so there is no repulsive coulomb force. and all that remains is the attractive magnetic force.

 

[Hornbein]

He's referring to Lenz's Law, where to wires with current in the same direction will attract one another. This is due to a relativistic effect. Due to their motion the electrons see more protons in the other wire than electrons. This of course does not occur with the two electron beams, which would repel one another.

 

[hutchphd]

He's referring to Lenz's Law, where to wires with current in the same direction will attract one another.

How is this illustrative of Lenz's Law??

 

[Hornbein]

How is this illustrative of Lenz's Law??

Oops it isn't. Shows how much I know. The attraction between two wires appears to have no particular name.

 

[tech99]

I believe that a trace of gas in the tube causes the beam to focus itself, presumably due to the introduction of positive ions.

 

[vanhees71]

Oops it isn't. Shows how much I know. The attraction between two wires appears to have no particular name.

It's the (magnetic) Lorentz force on the electrons in one wire moving in the magnetic field of the other wire.

Take wire 1 as fixed along the $z$-axis of our Cartesian coordinate system. Make it, for simplicitly, "infinitesimally thin". The current density then is
$$\vec{j}_1(\vec{r})=I_1 \delta(x) \delta(y) \vec{e}_z.$$
For symmetry reasons the magnetic field is given (in polar coordinates $(R,\varphi,z)$) by
$$\vec{B}(\vec{r})=B(R) \vec{e}_{\varphi}.$$
Using Ampere's Law with a circle around the $z$-axis, you get
$$2 \pi R B(R)=\mu_0 I_1 \Rightarrow \; B(R)=\frac{\mu_0 I_1}{2 \pi R} \vec{e}_{\varphi}.$$
Now take another infinitely thin wire parallel to the first one given by $x=d, \quad y=0, \quad z \in \mathbb{R}$. Then
$$\vec{j}_2=I_2 \delta(x-d) \delta(y) \vec{e}_z.$$
Then the force on a piece of length $L$ is
$$\vec{F} = \int_{\mathbb{R}} \mathrm{d} x \int_{\mathbb{R}} \mathrm{d} y \int_0^L \mathrm{d} z \vec{j}_2(\vec{r}) \times \vec{B}(\vec{r}) = \frac{\mu_0 I_1 I_2 L}{2 \pi d} \vec{e}_z \times \vec{e}_y=-\frac{\mu_0 I_1 I_2}{2 \pi d} \vec{e}_x.$$
This means that for $I_1 I_2>0$ (i.e., both current densities in the same direction) the wires are attracted, otherwise repelled.

There is no electric force, because both wires can be considered (for all practical purposes) as uncharged.