# Homework Help: Work per charge

1. Apr 11, 2006

### ak416

This is a question regarding electrodynamics. Specifically im reffering to p.294-296 of Griffiths 2rd edition. If you have the book it will be easier to see what my question is if you see the picture. There a loop with a resistor on the right end, part of the left side of the loop is in a magnetic field pointing into the page. The loop is pulled at velocity v to the right. This creates a magnetic force acting on the charges on the left end making them go up. Now to calculate the EMF i know from the formula that its the line integral around the whole loop of Fmag per charge . dl and that turns out to be vBh, where h is vertical height of loop. Now to calculate the work done per charge by the agent who is pulling the loop, they do an integration (using the two components of velocity and angles between horizontal pulling force) and it turns out to equal the EMF, and that makes sense. But by definition of work per charge, I know its the integral over the path of one charge that the force acts upon of Fpull . dl . But here's what confuses me. How do you know when the pulling force is acting on the charge or when the charge is free. Firstly, the force Fpull, acts indirectly, by pulling on the right end, so its kinda hard to see the direct connection. Also, once the charge reaches the top left edge (point b), is there still a pulling force acting on it? What about when it does back down the resistor? Cause if you still take the integral of Fpull . dl as it goes back down the resistor, this contributes to work of EMF also. So total work as the charge goes around the whole loop is 2 times EMF. So, basically, do I integrate over the whole loop when calculating the work done by the pulling force, or just over the little segment from a to b? On p. 295 they calculate just the little segment but on p. 296 they show a quite different path of integration which i dont understand...Please take a look guys if you can.

2. Apr 12, 2006

### Galileo

Hopefully I'm addressing the right problem, because the question is not entirely clear to me.

You only have to take into account the magnetic force acting on the charges in the left vertical segment of the loop, because the charges in the horizontal segments experience a force perpendicular to the wire and the rest is outside the magnetic field. The point of application of the pulling force is not important. It's a rigid loop and a force acting on it anywhere will accelerate it accordingly. You are not pulling directly on the charges, ofcourse, you are pulling the loop and the charges are constrained to stay in the loop, so the force is 'transmitted' to the charges by constraint. For the reason stated above, there is no contribution to the pulling force against charges in the horizontal segments (these do segments cause the loop to rotate, so you have to apply an additional force or torque on the loop to keep it from rotating, but no work is required for this since the force is perpendicular to the motion). Also, there is no contribution to the pulling force coming from the right horizontal segment, since the magnetic field there is zero. You only have to counteract the magnetic force in the left segment (ignoring the torque).

3. Apr 12, 2006

### ak416

so, in more detail, why isnt the pulling force contributing to 1)the top horizontal. 2)the right (not entirely vertical because the charges have 2 velocity components). 3)the bottom horizontal. segments? I thought you're supposed to integrate around the WHOLE loop and calculate F_pull . dl for each segment. So what you are saying is that F_pull=0 on these 3 segments? Or do the two horizontal ones cancel out? But then you would be left with a horizontal velocity component on the right segment and F_pull . dl wouldnt be zero there (unless F_pull is zero there).

4. Apr 13, 2006

### Galileo

If you want to know what force you need to apply to counteract the magnetic force isn't it obvious you only have to take into account the force acting perpendicular to the direction you are pulling? Nevermind WHERE the pulling force is applied or where it acts. It acts on the loop as a whole. The force is transmitted to the charges by the structure of the loop.

For the work done per unit charge you only have to consider the left vertical segment, since the motion is perpendicular to the motion in the horizontal parts and the magnetic field is zero at the right end.

5. Apr 14, 2006

### ak416

ok, im starting to get it now. But what about this: When you pull on the entire loop of charges, arent you intrinsically pulling on each individual charge (this includes the charges on the top, bottom and right segments)? I know that it is not necessary to pull on these because they will move with constant velocity (assuming no friction) since there are no magnetic fields opposing their motion. But using this logic, in computing the work done by the force F_pull on any given charge as it moves around the loop, i would have to compute F . dl on those 3 segments as well. So where am i wrong here?

6. Apr 15, 2006

### Galileo

Yes, these charges do not "stay behind" simply because there there would otherwise be a charge pile-up at the corner, so these charges in the horizontal segments get 'pushed' along. So the end result when pulling the loop is simply that all charges get velocity in the direction you are pulling.
You are really looking at one particular instant in time. You are not following a charge around the loop, you are calculating $\vec F \cdot d\vec l$ for each little charge in the loop. The sum of these contributions is the same as following a single charge around the loop.
So yes, in general you have to take into account the entire loop, but since Fpull is zero on the horizontal segments and in the right segment (where B=0) it's only the left part contributing.

If you have same weirdly shaped loop in a nonuniform B-field, then the analysis will be more difficult and each part of the loop will contribute, but the end result is the same.

7. Apr 15, 2006

### ak416

Didnt you just say yes to my question: "arent you intrinsically pulling on each individual charge (this includes the charges on the top, bottom and right segments)?"

Last edited: Apr 15, 2006
8. Apr 15, 2006

### ak416

The only way i could see how to resolve this would be to assume that when you pull on the loop, you're only pulling on the charges that need to be pulled (left segment) to balance the left magnetic force on them. But it seems kinda weird that nature would work that way, efficiently selecting to pull only those ones...

9. Apr 15, 2006

### Galileo

Yeah, Fmag is zero there, so you there's no contribution to Fpull from these segments. As before, it doesn't matter where Fpull is applied. I hope it's clear now.