Work done by Magnetic Forces

neelakash

magnetic forces never do work.

da_willem

The definition of work is integral of F.dr. If the magnetic force on an object is either parallel of antiparallel to the direction of motion, as it is in this case, there should be a nonzero work done by the magnetic force, and thus by the magnetic field. What is the explanation here?
The equally charged particles will be repulsed by the electric force, which does the work that supplies this kinetic energy. This can most easily be seen in an inertial frame moving along with the particles, where the situation is the extremely simple one of two equally charged particles repulsing each other by the normal Coulomb force.

When you switch to the 'nonmoving' frame, relative to which the particles move with a certain velocity a magnetic field 'arises' in our description. Note that for velocities smaller than c, this attraction force is always smaller than the electric repulsion force.

In this stationary frame the repulsion force is thus diminished by the magnetic field, but should ofcourse yield the same acceleration. I think this has something to do with time dilation, if anyone knows I'm quite interested.

neelakash

Originally Posted by lugita15
The definition of work is integral of F.dr. If the magnetic force on an object is either parallel of antiparallel to the direction of motion, as it is in this case, there should be a nonzero work done by the magnetic force, and thus by the magnetic field. What is the explanation here?
F_mag=q(vxB)
and F.dl=q(vxB).(v dt)

daniel_i_l

Gold Member
I didn't have time to read past the first few threads but I think that the answer is obvious:
Let's say that one wire has a current going through it and the other one doesn't, in this case there will be no force. Only when the current is switched on in the second wire do they move towards each other. This means that the source of the work is whatever is "pumping" the electrons around in the wires. Since that wasn't very clear I'll explain a different way:
Lets say we have a satellite orbiting around Earth, it's spinning around and around but the only force acting on it is gravity which is perpendicular to the motion so it isn't doing any work! How can that be? Well in this case the easy answer is that since the satellite's energy stays constant throughout the orbit no work has to be done on it by Earth.
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?

neelakash

Good!That is a direct proof that the battery does the work.
Here the wire gains a sidewise motion that calls for a KE.This is suppiled by the battery.

I do not understand the relevance of the analogy of the planetary motion.

lugita15

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.

da_willem

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.

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.

daniel_i_l

Gold Member
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.

neha_dodeja

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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