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I have difficulty understanding exercise 24 in this document:

Two parallel wires I and II that are near each other carry currents i and 3i both in the same direction. Compare the forces that the two wires exert on each other.

(a) Wire I exerts a stronger force on wire II than II exerts on I.

(b) Wire II exerts a stronger force on wire I than I exerts on II.

(c) The wires exert equal magnitude attractive forces on each other.

(d) The wires exert equal magnitude repulsive forces on each other.

(e) The wires exert no forces on each other.

I think - if you use $F_m=IlB\sin \alpha$, which is Force on electric wire in uniform magnetic field - that $F_{II}=IlB\sin \alpha = 3IlB\sin \alpha > IlB\sin \alpha =F_{I}$ So answer would be b), but how is it possible because you have Newton's third law( the forces should be equal, but does it apply here) and there is not any magnetic field here. So do I use this Lorentz's law or which law do I use?

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3 Answers

up vote 2 down vote accepted

The $B$ in these equations refers to the magnetic fields each individual wire creates. These fields are proportional to the individual wires' own currents.

In other words, the force of wire I on wire II is $F_{I \to II} =I_{II} \ell B_{I} \sin \alpha$. What is $B_{I}$, how does it relate to $I_I$, and how does it differ from $B_{II}$?

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So is it $F_{I \rightarrow II} = I_{II}lB_I sin \alpha = 3I_{I}l \frac{B_{II}}{3} sin \alpha = I_{I}lB_{II} sin \alpha = F_{II \rightarrow I}$? –  laovultai Nov 1 '12 at 13:59
    
So you can use this formula although they are not in any uniform magnetic field? –  laovultai Nov 1 '12 at 14:05
    
In this case, you can use it because the wires are parallel and equidistant from one another. The magnetic fields only change with distance (not with translation along the wires' axes or rotation), so it is, essentially, that every point on a given wire feels the same magnetic field. –  Muphrid Nov 1 '12 at 14:08
    
So would the answer be d)? –  laovultai Nov 1 '12 at 14:30
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All we've talked about to this point are the force magnitudes. What determines the force's direction (attractive vs. repulsive) in this problem? Isn't there a vector form of the Lorentz force law that would give directions to these forces? –  Muphrid Nov 1 '12 at 14:33
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So it seems like the easiest way to do this problem is to consider three things:

(1) Ampere's Law for a wire: $B= \frac{I\mu_0}{2\pi r}$

(2) Lorentz Force: $F=qv \times B$

(3) $qv = lq/t= lI \implies F= lI \times B $

Using (1) we can get the magnetic field that will act on each current/wire (a field generated by a current does not act on that same current; a wire sill not push itself).

Using (3) you can find a the force acting on each wire along with the directions of that force.

Good luck!

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Keep in mind the following:

The current in wire 1 is smaller but interacts with the stronger B field of wire 2.

The current in wire 2 is larger but interacts with the weaker B field of wire 1.

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