| Colin
Prue |
question: what will be the current in a circular conducting band of radius 1m and resistance 1W if the magnetic flux density (B) through it is changing at -1 T s-1 Rearrangement of Neumann's equation gives
hence EMF=pV I can get this far...that's the maths (pah!) done, but now i can't figure the physics: What does this EMF represent? i have a closed loop of conductor therefore i will never be able to take a measurement of pd (from Kirchhoff's second law) between any two points - pd=0V as loop is symmetrical. can anybody confirm that indeed: I=pA. i have difficulty understanding how such an EMF manifests itself so am not confident about this answer. sorry about once again tainting this maths board with physics but my teacher sometimes can't help me with physics olympiad stuff |
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| Matthew
Smith |
I wondered about exactly this sort of problem when I did A-level physics, and I've only recently (in 2nd-year university physics) found an explanation. It's sort of obvious that the answer must be as you've given, but it just doesn't seem to make sense in terms of potential difference. Have you learnt about the electric field vector, E ? It is the force acting on a unit charge at a particular point due to the electric field. It can be helpful to think of it as a more fundamental quantity than the potential. In fact, you can construct a potential based on it, by integrating it (hopefully you've covered this). For electrostatics, the integral of E around a closed loop is always zero, because the electric field is conservative. This allows us to construct a single-valued potential. However, this is no longer the case if we have time-varying fields. The electric field does not need to be conservative because energy can be exchanged through the magnetic field. This is clearest to see in the Maxwell equations, which are four vector calculus equations which are fundamental to electromagnetism. The relevant equation is curl E = - dB /dt. Don't worry about the quantity on the left, but it expresses, in some sense, the extent to which E isn't conservative. From this it can be seen that E is conservative only when B isn't changing with time. By applying a vector calculus theorem (called Stokes' theorem) to this equation, we get the result that The integral of E around a closed loop is equal to the integral of - dB /dt over the surface enclosed by the loop. By using the concept of magnetic flux, the latter integral becomes minus the rate of change of flux, and by regarding potential as the integral of E , the former becomes the E.M.F. acting 'round the loop'. Thus we recover Faraday's law of induction (Neumann's equation), which you are taught at A-level. However, this works even when the idea of potential seems meaningless, because all we need to do is work out E . What you have worked out as the 'E.M.F.' is the integral of E around the circle, and since, by symmetry, E is constant in magnitude, and we know it always points round the circle, we can get the value of E by dividing the 'E.M.F.' by the circumference of the circle. All we need now is a way of working out the current from E . The general equation is J = sigma E , where J is the current per unit area, and sigma is the conductivity (which is the reciprocal of the resistivity) of the metal. From this equation, and the value for E , you can get the current (as the length and cross sectional area of the wire will cancel out). Which should allow you to show, properly, that the current is indeed pi A. This seems rather a convoluted argument, but it makes much more sense when you've learnt about Maxwell's equations. The main thing to realise is that the fundamental quantity that you're dealing with is E , and that the E.M.F is just a convenient measure of it along a length, not a 'voltage'. It is the electric field that acts to create a current at each point. If you want more information, or better explanations, please post again, and I'll try. As a final disclaimer, I've made out in the above that a potential for the electric field only exists for statics, and someone's going to point out that this isn't the case. There is a more general concept of electric and magnetic potential which always works, but then the two are mixed up together, and it's all more complicated. |
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| Colin
Prue |
thanks a lot matthew. i think that i now understand (at some level at least!) what is happening. i have done some electric field questions that require fairly simple 2-variable line integrals, but for some reason i had never really made the connection between that and what goes on in a circuit! I feel slightly happier though that the answer that seemed most intuitive to me (albeit probably the only reasonable one) turned out to be the state of affairs thanks very much colin |