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.(adsbygoogle = window.adsbygoogle || []).push({});

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# Homework Help: Work per charge

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