Magnetic Forces and Work: An Explanation

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The discussion centers on the concept that a static magnetic field does no work on moving charged particles, as the magnetic force is always perpendicular to their motion. However, when considering two parallel currents, the magnetic field generated by one current can attract the other, leading to movement. This movement occurs because the deflection of charges in the wire induces an electric field, which ultimately does the work in moving the wires together. The participants clarify that while the magnetic field itself does not do work, it influences the motion of charges through the Lorentz force, which includes electric forces that do perform work. The conversation also touches on the complexities of magnetic interactions in quantum mechanics, particularly regarding magnetic dipoles and their behavior in magnetic fields.
  • #31
daniel_i_l said:
The same thing applies here. In order to find the source of the work done on the wires we have to look at the energy source of the system; who looses energy by the force moving the wires? The source of the current (batteries..) of course.
Am I missing something?
I believe you are. The cause of motion is not necessarily the force which does work. Work is done when and only when integral F.dr is 0.
 
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  • #32
lugita15 said:
I believe you are. The cause of motion is not necessarily the force which does work. Work is done when and only when integral F.dr is 0.

I think he says you have to look where the energy dissipates, and then look at the most direct force (with F.dr nonzero) that is repsonsible.
 
  • #33
lugita15 said:
I believe you are. The cause of motion is not necessarily the force which does work. Work is done when and only when integral F.dr is 0.

In response to your original post, the force is in fact perpendicular to the motion of the charge, leading to no work. I'll do my best to try and explain why. So I'll label everything as: wire 1, charge 1(travels in wire 1), wire 2, and charge 2(travels in wire 2). You said the direction of current is from left to right so I'll use the same convention. The X's stand for the magnetic field, and for wire 1, they are going into your computer screen between the wires and out of the screen on the other side.

OOOOOOOOOOOOOOOO
------------------------ wire 1
XXXXXXXXXXXXXXXXXX

OOOOOOOOOOOOOOOOO
------------------------ wire 2
XXXXXXXXXXXXXXXXXXX

So using the right hand rule, the force due to the field of wire 1(look at the field in between the lines, and it is into the page) on charge 2( moving from left to right in wire 2) will be directed from wire 2 towards wire 1(from the bottom to the top of the screen). The same idea holds for the field from wire 2 acting upon charge 1, which is directed from wire 1 to wire 2(from the top to bottom of the screen). Ok, so we've established the directions of the forces and now need to find the work done BY EACH MAGNETIC FIELD ON ITS RESPECTIVE CHARGE. So this means the work done by the field from wire 1 on charge 2 and the work done by the field from wire 2 on charge 1. The equation for work is W = Fdcos(theta). So we're interested in the parallel component of F stands for the force on the object(Ex. force from magnetic field of Wire 2 on charge 1 that goes from bottom to top of screen). However there will never be a parallel component of force on the object because the charge is traveling from left to right, thus making the distance traversed in some infitesimal amount of time also from left to right. Now due to the force, the charge may be deflected slightly towards the other wire, but at the same time, the force from the magnetic field is also going to change respectively, and it will still be oriented 90 degrees to the charge. So at all times, the force due to the magnetic field on a moving charge will always be zero.

Hope this helps.
 
  • #34
da_willem said:
I think he says you have to look where the energy dissipates, and then look at the most direct force (with F.dr nonzero) that is repsonsible.

Yes, that's basically what I meant.
 
  • #35
when we solve questions we consider the work done as ilb if the work is done by the electric field how come we have this expression??could u pls derive it for me.
 

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