1
$\begingroup$

Plastic Rod with Aluminium Foil

An aluminium strip is hung over a plastic rod. The ends P and R are connected by a battery and large current flows through the aluminium foil from P to R.

Why will the ends P and R repel each other? Or will it not?

I consider it as a simple circuit where the aluminium foil is the wire. While I know that, if there is currect flows through the wire, there will be magnetic field, but can that magnetic field affect the wire itself ?

So, if I bend the any wire into U-shaped, the wire will attract to itself ?

$\endgroup$
3
  • $\begingroup$ Why would it not affect it? $\endgroup$
    – biziclop
    Commented Mar 16, 2017 at 15:28
  • $\begingroup$ @biziclop When we are doing experiment using electric circuit, we don't see the wires attract or repel each other. So, I take it as evidence that it wouldn't. I may be naive though. $\endgroup$ Commented Mar 16, 2017 at 15:39
  • $\begingroup$ Here's a video showing the effect. youtube.com/watch?v=43AeuDvWc0k $\endgroup$
    – David Elm
    Commented Mar 16, 2017 at 16:24

3 Answers 3

11
$\begingroup$

Your question is a classic question. Yes the 2 wires will repel each other. The classic explanation is that this is the result of the $\vec{v}\times\vec{B}$ Lorentz force. Here $\vec{B}$ is the magnetic field that is produce by the current (Ampere's Law). This force is in opposite direction in the 2 aluminum "wires". Now comes the 64 million \$ question. We have learned that $q\vec{v}\times\vec{B}$ force cannot do any work as this force is perpendicular to $\vec{v}$. The answer is by no means obvious. The work is done by electric forces.

Here follows a note from my former colleague Professor Barton Zwieback at MIT.

Walter The issue of work by magnetic fields usually comes up. I have not yet looked at your example in detail, but the typical resolution is that the magnetic force never does work but allows other agent to do it.

This is explained in Griffiths’s Intro to Electrodynamics book, pages 210-211.

There are interesting analogs of this. Suppose you have a frictionless ramp and you push an object horizontally against it. The object will go up. The only force with up component acting on the object is the normal force perpendicular to the ramp. So it would seem that the normal force is pushing the object up. This is true but the normal force does not do work. The work is done by you, pushing horizontally. The normal force helps channel the motion upwards.

$\endgroup$
0
3
$\begingroup$

You don't usually see the effect of the magnetic force between two current-carrying wires because this force is quite weak. In demonstrations it is usual to use long strips of aluminium foil loosely suspended, because it is light and easily moved - eg by a breath of air. If the foils or wires were resting on a table, the force of friction would probably be too great for the magnetic force to overcome.

The force exists between any two current-carrying elements; they do not have to be part of different wires. If the wire or foil forms a loop the magnetic force repels (not attracts) parallel sections which carry opposite currents. The resulting forces try to expand the loop outwards. But again the force is weak and might have no effect because of friction.

In your case the weight or stiffness of the leads attached to the ends of the foil might prevent them from moving a noticeable distance.

$\endgroup$
3
$\begingroup$

This type of behavior (current in parallel sections of a conductor) is used to make the current (!!) official definition of the ampere:

One ampere is that unvarying current that, if present in each of two parallel conductors of infinite length and one meter apart in empty space, causes each conductor to experience a force of exactly $2\times 10^{-7}$ newtons per meter of length. (Young & Freedman, 9th edition, page 912)

If the currents in parallel wires are in opposite directions there will be repelling forces.

Current balances work from this same principle.

$\endgroup$

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