swansont Posted March 16, 2012 Share Posted March 16, 2012 I have only I, not i. If you consider another current i, which is different, you are not talking about my circuit. You have an EMF, don't you? From the motion of the bar? Then you have a current from that EMF. If you set the problem up so that the current source is always giving you a net current, above and beyond that provided by the EMF, then we're back to requiring an external force to keep the velocity at v, and that external force does work on the system. Link to comment Share on other sites More sharing options...
pengkuan Posted March 16, 2012 Author Share Posted March 16, 2012 (edited) You have an EMF, don't you? From the motion of the bar? Then you have a current from that EMF. I have said with a link Wikipedia In circuit theory, an ideal current source is a circuit element where the current through it is independent of the voltage across it. You know it is a current source but you still make your current vary. It is wrong If you set the problem up so that the current source is always giving you a net current, above and beyond that provided by the EMF, then we're back to requiring an external force to keep the velocity at v, and that external force does work on the system. Yes there is external electric work, dW(emf) = I U dt= I l B dx, in my post. Edited March 16, 2012 by pengkuan Link to comment Share on other sites More sharing options...
swansont Posted March 16, 2012 Share Posted March 16, 2012 I have said with a link Wikipedia In circuit theory, an ideal current source is a circuit element where the current through it is independent of the voltage across it. You know it is a current source but you still make your current vary. It is wrong A current source puts out a constant current (I), but that does not mean the current in the rest of the circuit will be the same, if you have another source. In this case, from the EMF, giving you a current of -i. Yes there is external electric work, dW(emf) = I U dt= I l B dx, in my post. The current from the EMF is not the same as the current from the current source; we've established that one is constant and the other varies with speed. In any event, this is not external to the system, it is internal, and the EMF does not exert a force. An external force is required for the bar to move at constant v. Link to comment Share on other sites More sharing options...
pengkuan Posted March 16, 2012 Author Share Posted March 16, 2012 A current source puts out a constant current (I), but that does not mean the current in the rest of the circuit will be the same, if you have another source. In this case, from the EMF, giving you a current of -i. You mean in the same loop, the current in the current source is different from that in the bar? The current from the EMF is not the same as the current from the current source; we've established that one is constant and the other varies with speed. In any event, this is not external to the system, it is internal, and the EMF does not exert a force. An external force is required for the bar to move at constant v. Yes, you really think "The current from the EMF is not the same as the current from the current source; ". They are in series. Link to comment Share on other sites More sharing options...
swansont Posted March 16, 2012 Share Posted March 16, 2012 You mean in the same loop, the current in the current source is different from that in the bar? If there is an induced EMF, yes. Yes, you really think "The current from the EMF is not the same as the current from the current source; ". They are in series. And in opposite directions, in accordance with Lenz's law. We've already established that there is no EMF when v=0, haven't we? There is no way the induced EMF can generate a current in that case. Remember, you're the one who thinks this violates conservation of energy and nobody noticed it for >100 years. Most people would simply recognize that they had erred in the solution to the problem. Link to comment Share on other sites More sharing options...
pengkuan Posted March 16, 2012 Author Share Posted March 16, 2012 If there is an induced EMF, yes. The current is imposed by the current source. That is, the loop that passes through it has the same current what ever it happens. So, in the bar the current is I and is independent of v, even there is an emf. Remember, you're the one who thinks this violates conservation of energy and nobody noticed it for >100 years. Most people would simply recognize that they had erred in the solution to the problem. This is the point. Link to comment Share on other sites More sharing options...
swansont Posted March 16, 2012 Share Posted March 16, 2012 The current is imposed by the current source. That is, the loop that passes through it has the same current what ever it happens. So, in the bar the current is I and is independent of v, even there is an emf. And the current created by the EMF? What happens to that? This is the point. Indeed. Meanwhile, motors actually work. Link to comment Share on other sites More sharing options...
pengkuan Posted March 16, 2012 Author Share Posted March 16, 2012 And the current created by the EMF? What happens to that? compensated by the current source. Indeed. Meanwhile, motors actually work. yes. The change is in the calculation of motors. Link to comment Share on other sites More sharing options...
swansont Posted March 16, 2012 Share Posted March 16, 2012 compensated by the current source. OK, that's not a standard use of a current source, but fine. The source must put out I + i, where i is the current from the EMF, such that you always have a current I in the loop. Then you have a Lorentz force on the wire, but in your setup there is still no other force. Yet it moves with constant v. You need an external force. yes. The change is in the calculation of motors. The calculation works, too. If you do it correctly. Link to comment Share on other sites More sharing options...
pengkuan Posted March 16, 2012 Author Share Posted March 16, 2012 OK, that's not a standard use of a current source, but fine. The source must put out I + i, where i is the current from the EMF, such that you always have a current I in the loop. Then you have a Lorentz force on the wire, but in your setup there is still no other force. Yet it moves with constant v. You need an external force. The work of external force is in my post this: dW(mec) = F dx= I l B dx (I have done a typo, it is dW(mec) here, not dW(emf)) The calculation works, too. If you do it correctly. yes Link to comment Share on other sites More sharing options...
swansont Posted March 17, 2012 Share Posted March 17, 2012 The work of external force is in my post this: dW(mec) = F dx= I l B dx (I have done a typo, it is dW(mec) here, not dW(emf)) yes OK, so do you still think there is a problem? Link to comment Share on other sites More sharing options...
pengkuan Posted March 17, 2012 Author Share Posted March 17, 2012 OK, so do you still think there is a problem? yes Link to comment Share on other sites More sharing options...
swansont Posted March 17, 2012 Share Posted March 17, 2012 yes Why? How much net energy is the current source dumping into the system? Link to comment Share on other sites More sharing options...
pengkuan Posted March 17, 2012 Author Share Posted March 17, 2012 The current source does work dW(emf) = I U(emf) dt= I l B dx, The Lorentz force does work dW(mec) = F(Lorentz) dx= I l B dx They are equal. I l B dx=I l B dx, dW(emf) = dW(mec) Why? How much net energy is the current source dumping into the system? see above Link to comment Share on other sites More sharing options...
swansont Posted March 17, 2012 Share Posted March 17, 2012 The current source does work dW(emf) = I U(emf) dt= I l B dx, The Lorentz force does work dW(mec) = F(Lorentz) dx= I l B dx They are equal. I l B dx=I l B dx, dW(emf) = dW(mec) see above No, the current from the EMF is not I, it is i (and is a function of the speed, v), and since it has nothing to do with the work done by the Lorentz force you can;t just set them equal. You said this was a typo earlier, but have simply written the same thing down. That's a mistake. Because of the EMF, the current source has to supply a current I+i to the circuit, in order for a current I to exist in the loop. The work done by the Lorentz force is balanced by the mechanical, but is only due to I. Your setup requires an additional current i from the current source, adding energy to the system. You're missing that. Do a proper analysis, and you will find that energy is conserved. Link to comment Share on other sites More sharing options...
pengkuan Posted March 17, 2012 Author Share Posted March 17, 2012 No, the current from the EMF is not I, it is i (and is a function of the speed, v), and since it has nothing to do with the work done by the Lorentz force you can;t just set them equal. You said this was a typo earlier, but have simply written the same thing down. That's a mistake. Because of the EMF, the current source has to supply a current I+i to the circuit, in order for a current I to exist in the loop. The work done by the Lorentz force is balanced by the mechanical, but is only due to I. Your setup requires an additional current i from the current source, adding energy to the system. You're missing that. Do a proper analysis, and you will find that energy is conserved. We do not have the same physics. For me, the current is I, imposed, the voltage between its 2 electrodes is U. The power is I U. During time dt, the energy is I U dt. In the loop, the curent is I, the voltage is EMF. So, the energy is dW(emf)=I.U(emf) dt. Link to comment Share on other sites More sharing options...
swansont Posted March 17, 2012 Share Posted March 17, 2012 We do not have the same physics. That's quite obvious. For me, the current is I, imposed, the voltage between its 2 electrodes is U. The power is I U. During time dt, the energy is I U dt. In the loop, the curent is I, the voltage is EMF. So, the energy is dW(emf)=I.U(emf) dt. And you would be wrong. You agreed that the EMF is zero when v=0. Your solution fails. Link to comment Share on other sites More sharing options...
pengkuan Posted March 17, 2012 Author Share Posted March 17, 2012 That's quite obvious. And you would be wrong. You agreed that the EMF is zero when v=0. Your solution fails. In my physics , we use 2 types of ideal electric source, voltage source and current source . voltage source: tension U0 does not change. deliver 1 A, 2A 100A ... with the same tension. current source: current I0 constant. deliver 1V, 2V 100V .... with the same current. You use a "current source", that deliver I, say 1A, then, with tension 1V (due to emf), the current becomes 1+0.1A, if the tension is 10V (emf), the current is 1+10*0.1=2A This is a current varying source. It is of unknown kind. you are talking about another circuit than me. The mine does not conserve energy. The your, I don't know. Link to comment Share on other sites More sharing options...
swansont Posted March 18, 2012 Share Posted March 18, 2012 In my physics , we use 2 types of ideal electric source, voltage source and current source . voltage source: tension U0 does not change. deliver 1 A, 2A 100A ... with the same tension. current source: current I0 constant. deliver 1V, 2V 100V .... with the same current. You use a "current source", that deliver I, say 1A, then, with tension 1V (due to emf), the current becomes 1+0.1A, if the tension is 10V (emf), the current is 1+10*0.1=2A This is a current varying source. It is of unknown kind. you are talking about another circuit than me. The mine does not conserve energy. The your, I don't know. You are contradicting yourself. If the current source delivers 1A but there is a back EMF, there will be less than 1A in the circuit. But you specifically stated that the current in the circuit is a constant, which means the source has to deliver more than 1A. So, which is it? Either way, though, you are in error. There is no conundrum here if you properly analyze the circuit. Link to comment Share on other sites More sharing options...
pengkuan Posted March 18, 2012 Author Share Posted March 18, 2012 You are contradicting yourself. If the current source delivers 1A but there is a back EMF, there will be less than 1A in the circuit. But you specifically stated that the current in the circuit is a constant, which means the source has to deliver more than 1A. So, which is it? Either way, though, you are in error. There is no conundrum here if you properly analyze the circuit. you win but the paradox stays Link to comment Share on other sites More sharing options...
swansont Posted March 18, 2012 Share Posted March 18, 2012 you win but the paradox stays No. If you analyze it correctly there is no paradox. And when you analyze it incorrectly, there should be no surprise that you find an inconsistency with a valid analysis. Link to comment Share on other sites More sharing options...
pengkuan Posted March 18, 2012 Author Share Posted March 18, 2012 Please read what you said: If you analyze it correctly ..... And when you analyze it incorrectly, .... ... That's a mistake. ... Do a proper analysis, .... OK, that's not a standard use of a current source, ........... If you do it correctly. .... And you would be wrong. .... Your solution fails. You are contradicting yourself. ..... Either way, though, you are in error. ..... if you properly analyze the circuit. your objective is to win. Look at how you speak in previous posts. So , let it be. You win. But you did not solve the paradox, because you are thinking in term of voltage source plus resistor. Only in this way, the current can vary with an additional tension. I have said all the way that the current is imposed by the current source. You did not listen to me at all and insisted on using voltage source. So you do not analyse the same circuit than me. It does not solve my circuit's paradox for sure. Link to comment Share on other sites More sharing options...
swansont Posted March 18, 2012 Share Posted March 18, 2012 your objective is to win. Look at how you speak in previous posts. So , let it be. You win. My objective is to convey correct science and correct flawed science. Your claim of a paradox is false. But you did not solve the paradox, because you are thinking in term of voltage source plus resistor. Only in this way, the current can vary with an additional tension. I have said all the way that the current is imposed by the current source. You did not listen to me at all and insisted on using voltage source. So you do not analyse the same circuit than me. It does not solve my circuit's paradox for sure. Either way you set it up, there is no paradox, unless you use a current source that defies the laws of physics. Then you can get any answer you want, and yes, it will break conservation of energy and continuity of current. You simply can't deliver a current I to the system and subtract a current i from a back EMF, and still have a current I in the loop. That's just blatantly wrong. Link to comment Share on other sites More sharing options...
pengkuan Posted March 18, 2012 Author Share Posted March 18, 2012 My objective is to convey correct science and correct flawed science. Your claim of a paradox is false. Either way you set it up, there is no paradox, unless you use a current source that defies the laws of physics. Then you can get any answer you want, and yes, it will break conservation of energy and continuity of current. You simply can't deliver a current I to the system and subtract a current i from a back EMF, and still have a current I in the loop. That's just blatantly wrong. thanks Link to comment Share on other sites More sharing options...
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