1
$\begingroup$

Now, I understand that when a an electron travels, it creates a magnetic field. If you put two wires with current traveling in the same direction they repel, and current traveling in opposite directions attract. I (somewhat) understand how this is caused by electrostatics and special relativity, but when I tried to get a better idea of the forces at work, I ended up with a result that said the force of attraction between to opposite wires was stronger than the force of repulsion between to wires with the same direction of current. Is this correct, or did I just mess up somewhere?

$\endgroup$
  • $\begingroup$ What is your better idea and how you ended up with the result? $\endgroup$ – HolgerFiedler Jul 15 '14 at 16:27
0
$\begingroup$

I would leave special relativity out of the picture. I mean: magnetism is a relativistic effect, in the sense that it pops out from the application of (special) relativity to electrostatics, but their relationship is more conceptual and may deserve a new question entirely devoted to the matter. Long story short, you don't need special relativity to understand magnetic forces.

All you need to know is that when a charged particle $q$ moves at a velocity $\mathbf{v}$ through an electric field $\mathbf{E}$ and a magnetic field $\mathbf{B}$, it experiences the Lorentz force $$ \mathbf{F_L} = q( \mathbf{E} + \mathbf{v}\times\mathbf{B}) $$

This really is the fundamental equation for electrodynamics, together with Maxwell's equations.

Now, a current-carrying wire is usually considered to be neutral overall (i.e. the number of electrons and ions are the same so that the net charge is zero), so you can take the electric field $\mathbf{E} = 0$.

Then, you are dealing with a current, not with a single charge. How to get the above equation in terms of the current $I$?

The electric current $I$ is defined to be $\frac{dq}{dt}$ where $q$ is the electric charge. The velocity is defined as $\frac{d\mathbf{l}}{dt}$ where $d\mathbf{l}$ is the path element in the direction of the current (or of the charge's trajectory).

So, the infinitesimal force on an infinitesimal charge $dq$ is $d\mathbf{F_L} = dq( \mathbf{v}\times\mathbf{B}) $, where $$ dq\cdot\mathbf{v} = dq\cdot\frac{d\mathbf{l}}{dt} = \frac{dq}{dt} d\mathbf{l} = Id\mathbf{l}$$.

To get the total force, you just integrate along the path defined by $d\mathbf{l}$:

$$\mathbf{F_L} = \int_L I d\mathbf{l}\times\mathbf{B} $$

OK. But you have two wires. And you know that a current carrying wire generates a magnetic field, given by (assuming the wire is in a straigh line) : $$ B_{\phi} = \frac{\mu_0}{2\pi}\frac{I}{r} $$ meaning that it is only in the tangential ($\neq$ radial) direction. The direction of the field is given by the right hand rule, line up your right thumb with the current direction and your fingers will tell you the direction of the magnetic field. This formula is obtained via Ampère's law.

If the two wires are in a straight line and are separated by a distance $d$, the value of the magnetic field due to one at the position of the other is $B = \frac{\mu_0}{2\pi}\frac{I}{d}$. Assume the wires carry equal, constant currents $I$.

Using this as the magnetic field in the Lorentz force, you get a force of $$ |\mathbf{F}| = I\cdot(\frac{\mu_0}{2\pi}\frac{I}{d})\cdot \int_L dl = \frac{\mu_0}{2\pi}\frac{I^2}{d}\cdot L $$ where $L$ is the length of the wires, or a force per unit length of $$ |\mathbf{F}| = \frac{\mu_0}{2\pi}\frac{I^2}{d} $$.

$d\mathbf{l}$ points along the current: use this and the cross product in the Lorentz force equation to get the direction of the resulting force. If the currents are parallel, the force is attractive, while if they are opposite, the force is repulsive. The magnitudes of the forces are the same, as they only depend on the wire separation $d$ and on the current $I$.

Notice, the other wire will exert the exact same force on the fist, due to Newton's III law of motion.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.