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Thread: Magnetic induction

  1. #1 Magnetic induction 
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    Hi!
    In my physics textbook there is a problem which reads as follows:
    A quadratic wire loop is placed in a homogenous magnetic field of absolute field strength B. The wire is moved to the right (in the picture not shown here), and initially the entire wire loop is inside the field, but after some time time the right edge leaves the field. Will induction in the wire loop occur?

    I can easily solve this problem by using Faraday's law of induction. In the initial phase there is no change of flux, hence not voltage induced, according to previously mentioned law. When the right edge of the loop leaves the field however, the area "exposed" to the field will decrease, hence a voltage will be induced by Faraday's law of induction.

    However, I also thought of it this way. Consider two metal rods, or pieces of wire, of equal length, placed parallell in a magnetic field and moving with the same speed perpendicularly to the magnetic field. In each rod there will be a voltage induced according to e=vBl, where e is the induced voltage, v is the velocity relative to the field, B is the magnetic flux density and l is the length of the rods.
    For the sake of argument, let's say that the lower part of each rod gets negative potential relative to the upper part of same rod. If I now conenct the facing ends of the two parallell rods at both sides, I will connect points of the same potential and thus not change anything. So there still must be a voltage induced, but now we have a wire loop, in which we have already shown, no voltage will be induced.

    Could somebody explain what is wrong in my argument, please?


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  3. #2 Re: Magnetic induction 
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    Quote Originally Posted by thyristor
    Hi!
    In my physics textbook there is a problem which reads as follows:
    A quadratic wire loop is placed in a homogenous magnetic field of absolute field strength B. The wire is moved to the right (in the picture not shown here), and initially the entire wire loop is inside the field, but after some time time the right edge leaves the field. Will induction in the wire loop occur?

    I can easily solve this problem by using Faraday's law of induction. In the initial phase there is no change of flux, hence not voltage induced, according to previously mentioned law. When the right edge of the loop leaves the field however, the area "exposed" to the field will decrease, hence a voltage will be induced by Faraday's law of induction.

    However, I also thought of it this way. Consider two metal rods, or pieces of wire, of equal length, placed parallell in a magnetic field and moving with the same speed perpendicularly to the magnetic field. In each rod there will be a voltage induced according to e=vBl, where e is the induced voltage, v is the velocity relative to the field, B is the magnetic flux density and l is the length of the rods.
    For the sake of argument, let's say that the lower part of each rod gets negative potential relative to the upper part of same rod. If I now conenct the facing ends of the two parallell rods at both sides, I will connect points of the same potential and thus not change anything. So there still must be a voltage induced, but now we have a wire loop, in which we have already shown, no voltage will be induced.

    Could somebody explain what is wrong in my argument, please?
    Your argument is fine. When you connect the two rods as you suggest there is zero voltage change around the loop. There will be no current flow.


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  4. #3 Re: Magnetic induction 
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    Quote Originally Posted by DrRocket
    Quote Originally Posted by thyristor
    Hi!
    In my physics textbook there is a problem which reads as follows:
    A quadratic wire loop is placed in a homogenous magnetic field of absolute field strength B. The wire is moved to the right (in the picture not shown here), and initially the entire wire loop is inside the field, but after some time time the right edge leaves the field. Will induction in the wire loop occur?

    I can easily solve this problem by using Faraday's law of induction. In the initial phase there is no change of flux, hence not voltage induced, according to previously mentioned law. When the right edge of the loop leaves the field however, the area "exposed" to the field will decrease, hence a voltage will be induced by Faraday's law of induction.

    However, I also thought of it this way. Consider two metal rods, or pieces of wire, of equal length, placed parallell in a magnetic field and moving with the same speed perpendicularly to the magnetic field. In each rod there will be a voltage induced according to e=vBl, where e is the induced voltage, v is the velocity relative to the field, B is the magnetic flux density and l is the length of the rods.
    For the sake of argument, let's say that the lower part of each rod gets negative potential relative to the upper part of same rod. If I now conenct the facing ends of the two parallell rods at both sides, I will connect points of the same potential and thus not change anything. So there still must be a voltage induced, but now we have a wire loop, in which we have already shown, no voltage will be induced.

    Could somebody explain what is wrong in my argument, please?
    Your argument is fine. When you connect the two rods as you suggest there is zero voltage change around the loop. There will be no current flow.
    Okay, but doesn't this conflict with the argument deduced from Faraday's law, then?
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    I thought you said one edge of the loop left the magnetic field.
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    Quote Originally Posted by Harold14370
    I thought you said one edge of the loop left the magnetic field.
    That's right, but not initially. My point was that I can understand why there is a voltage induced when one edge leaves the field, both by using Faraday's law of induction and my own argument. But when both edges are still in the field, Faraday's law of induction tells me no voltage is induced, but my argument tells me that a voltage is, in fact, induced.
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    Quote Originally Posted by thyristor
    Quote Originally Posted by Harold14370
    I thought you said one edge of the loop left the magnetic field.
    That's right, but not initially. My point was that I can understand why there is a voltage induced when one edge leaves the field, both by using Faraday's law of induction and my own argument. But when both edges are still in the field, Faraday's law of induction tells me no voltage is induced, but my argument tells me that a voltage is, in fact, induced.
    What argument tells you there is a voltage induced? If you connect two ends of a wire, each end being the same potential as the corresponding end of the other wire, no current will flow. Okay, there's voltage on the wires, from one side to another, but no voltage driving a current around the loop.
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    Quote Originally Posted by Harold14370
    Quote Originally Posted by thyristor
    Quote Originally Posted by Harold14370
    I thought you said one edge of the loop left the magnetic field.
    That's right, but not initially. My point was that I can understand why there is a voltage induced when one edge leaves the field, both by using Faraday's law of induction and my own argument. But when both edges are still in the field, Faraday's law of induction tells me no voltage is induced, but my argument tells me that a voltage is, in fact, induced.
    What argument tells you there is a voltage induced? If you connect two ends of a wire, each end being the same potential as the corresponding end of the other wire, no current will flow. Okay, there's voltage on the wires, from one side to another, but no voltage driving a current around the loop.
    Sure, I agree there's no current. But what does one then mean saying there is a voltage induced in a current loop? Between which points are we measuring the potential difference?
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    If you opened up the loop and inserted a voltmeter, you would measure a voltage. Or, if you put an ammeter probe around the wire you would detect a current flowing.
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  10. #9  
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    Quote Originally Posted by thyristor
    Sure, I agree there's no current. But what does one then mean saying there is a voltage induced in a current loop? Between which points are we measuring the potential difference?
    It means thatthe field is non-conservative -- the line integral around a closed loop is not zero.

    There is no potential field. If there were the field would be conservative.
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  11. #10  
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    Quote Originally Posted by Harold14370
    If you opened up the loop and inserted a voltmeter, you would measure a voltage. Or, if you put an ammeter probe around the wire you would detect a current flowing.
    But didn't we agree there was no current during the initial phase?
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  12. #11  
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    Quote Originally Posted by DrRocket
    Quote Originally Posted by thyristor
    Sure, I agree there's no current. But what does one then mean saying there is a voltage induced in a current loop? Between which points are we measuring the potential difference?
    It means thatthe field is non-conservative -- the line integral around a closed loop is not zero.

    There is no potential field. If there were the field would be conservative.
    I am afriad I am not yet familiar with line integrals.
    You say that there is no potential field, which is in contrast to what I deduced, so where does my argument fail?
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  13. #12  
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    Quote Originally Posted by thyristor
    Quote Originally Posted by DrRocket
    Quote Originally Posted by thyristor
    Sure, I agree there's no current. But what does one then mean saying there is a voltage induced in a current loop? Between which points are we measuring the potential difference?
    It means that the field is non-conservative -- the line integral around a closed loop is not zero.

    There is no potential field. If there were the field would be conservative.
    I am afriad I am not yet familiar with line integrals.
    You say that there is no potential field, which is in contrast to what I deduced, so where does my argument fail?
    You fail to understand the meaning of "potential" -- in this case a scalar field the gradient of which would be the E-field. Only conservative fields have potentials.

    There is a generalization, vector potentials, that applies to some non-conservative fields.

    If you have not yet studied line integrals, you are in over your head.
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  14. #13  
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    Quote Originally Posted by DrRocket
    Quote Originally Posted by thyristor
    Quote Originally Posted by DrRocket
    Quote Originally Posted by thyristor
    Sure, I agree there's no current. But what does one then mean saying there is a voltage induced in a current loop? Between which points are we measuring the potential difference?
    It means that the field is non-conservative -- the line integral around a closed loop is not zero.

    There is no potential field. If there were the field would be conservative.
    I am afriad I am not yet familiar with line integrals.
    You say that there is no potential field, which is in contrast to what I deduced, so where does my argument fail?
    You fail to understand the meaning of "potential" -- in this case a scalar field the gradient of which would be the E-field. Only conservative fields have potentials.

    There is a generalization, vector potentials, that applies to some non-conservative fields.

    If you have not yet studied line integrals, you are in over your head.
    Very well then. So let me rephrase my question: Suppose I opened up the wire loop between "the upper part" and the "loer part" during the "initial phase" (as introduced above) and connected a voltmeter. Would the instrument detect anything?
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  15. #14  
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    Quote Originally Posted by thyristor
    Quote Originally Posted by DrRocket
    Quote Originally Posted by thyristor
    Quote Originally Posted by DrRocket
    Quote Originally Posted by thyristor
    Sure, I agree there's no current. But what does one then mean saying there is a voltage induced in a current loop? Between which points are we measuring the potential difference?
    It means that the field is non-conservative -- the line integral around a closed loop is not zero.

    There is no potential field. If there were the field would be conservative.
    I am afriad I am not yet familiar with line integrals.
    You say that there is no potential field, which is in contrast to what I deduced, so where does my argument fail?
    You fail to understand the meaning of "potential" -- in this case a scalar field the gradient of which would be the E-field. Only conservative fields have potentials.

    There is a generalization, vector potentials, that applies to some non-conservative fields.

    If you have not yet studied line integrals, you are in over your head.
    Very well then. So let me rephrase my question: Suppose I opened up the wire loop between "the upper part" and the "loer part" during the "initial phase" (as introduced above) and connected a voltmeter. Would the instrument detect anything?
    If the wires were conductive, yes. If not, no. The reason for the "yes", is due to movement of electrons in the conductor due to the Lorentz force, and therefore due to the electric field that results from the unbalanced charge distribution. It does not come from a straightforward application of Faraday's law to the coupled magnetic field. This is related to the theory of the homopolar generator.

    You shlould read about induction and Faraday's Law in The Feynman lectures on Physics. There is thread, think it was finally mercifully confined to Pseudoscience, on Faraday's law that bears on your question. It can be subtle, but it is fully explained by classical electrodynamics.

    The basic problem here is that you are thinking deeply about a topic in electrodynamics which you do not yet have the mathematics to fully comprehend. Be patient. You will get there.
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  16. #15  
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    Quote Originally Posted by DrRocket
    Quote Originally Posted by thyristor
    Quote Originally Posted by DrRocket
    Quote Originally Posted by thyristor
    Quote Originally Posted by DrRocket
    Quote Originally Posted by thyristor
    Sure, I agree there's no current. But what does one then mean saying there is a voltage induced in a current loop? Between which points are we measuring the potential difference?
    It means that the field is non-conservative -- the line integral around a closed loop is not zero.

    There is no potential field. If there were the field would be conservative.
    I am afriad I am not yet familiar with line integrals.
    You say that there is no potential field, which is in contrast to what I deduced, so where does my argument fail?
    You fail to understand the meaning of "potential" -- in this case a scalar field the gradient of which would be the E-field. Only conservative fields have potentials.

    There is a generalization, vector potentials, that applies to some non-conservative fields.

    If you have not yet studied line integrals, you are in over your head.
    Very well then. So let me rephrase my question: Suppose I opened up the wire loop between "the upper part" and the "loer part" during the "initial phase" (as introduced above) and connected a voltmeter. Would the instrument detect anything?
    If the wires were conductive, yes. If not, no. The reason for the "yes", is due to movement of electrons in the conductor due to the Lorentz force, and therefore due to the electric field that results from the unbalanced charge distribution. It does not come from a straightforward application of Faraday's law to the coupled magnetic field. This is related to the theory of the homopolar generator.

    You shlould read about induction and Faraday's Law in The Feynman lectures on Physics. There is thread, think it was finally mercifully confined to Pseudoscience, on Faraday's law that bears on your question. It can be subtle, but it is fully explained by classical electrodynamics.

    The basic problem here is that you are thinking deeply about a topic in electrodynamics which you do not yet have the mathematics to fully comprehend. Be patient. You will get there.
    Okay, thanks for your answer. I guess I'm gonna have to read some more on the topic and related fields in mathematics.

    However I would like to ask a follow up question. Suppose the wires were super conductive. Applying the Lorentz force on the electrons in the wire, I can understand that they will move. But will there still be voltage? I reckon that in a superconductor, since there is mathematically zero resistance, any to points will have the same potential.
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  17. #16  
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    Quote Originally Posted by thyristor
    However I would like to ask a follow up question. Suppose the wires were super conductive. Applying the Lorentz force on the electrons in the wire, I can understand that they will move. But will there still be voltage? I reckon that in a superconductor, since there is mathematically zero resistance, any to points will have the same potential.
    There will be a charge imbalance between the ends of the superconductor, and an E-field comes with that. The E-field inside the conductor will be zero, once the transients have settled out. This does not depend on superconductivity, at least in the steady state. Resistance is not relevant since there is no current flow.
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  18. #17  
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    Quote Originally Posted by DrRocket
    Quote Originally Posted by thyristor
    However I would like to ask a follow up question. Suppose the wires were super conductive. Applying the Lorentz force on the electrons in the wire, I can understand that they will move. But will there still be voltage? I reckon that in a superconductor, since there is mathematically zero resistance, any to points will have the same potential.
    There will be a charge imbalance between the ends of the superconductor, and an E-field comes with that. The E-field inside the conductor will be zero, once the transients have settled out. This does not depend on superconductivity, at least in the steady state. Resistance is not relevant since there is no current flow.
    Allright, but consider a case where there is a current flow (say a superconductive wire loop, moving in a non-homogenous field). Would it then be possible to talk about voltage induced in the wire loop?
    I am sorry if my questions may seem stupid, but a lot of simplifications are made in our high school physics course, and due to lack of proper explanation in our education, I feel I do not fully grasp the concepts.
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  19. #18  
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    Quote Originally Posted by thyristor
    Allright, but consider a case where there is a current flow (say a superconductive wire loop, moving in a non-homogenous field). Would it then be possible to talk about voltage induced in the wire loop?
    I am sorry if my questions may seem stupid, but a lot of simplifications are made in our high school physics course, and due to lack of proper explanation in our education, I feel I do not fully grasp the concepts.
    You can talk about an EMF, but not a potential function. We are right bck to a non-conservative E-field and the line integral around a closed loop not being zero.

    The basic problem is that high school mathematics does not permit an adequate description of the physics.

    The only way that you are going to understand this is via a university-level treatment of electrodynamics.
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  20. #19  
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    Quote Originally Posted by DrRocket
    Quote Originally Posted by thyristor
    Allright, but consider a case where there is a current flow (say a superconductive wire loop, moving in a non-homogenous field). Would it then be possible to talk about voltage induced in the wire loop?
    I am sorry if my questions may seem stupid, but a lot of simplifications are made in our high school physics course, and due to lack of proper explanation in our education, I feel I do not fully grasp the concepts.
    You can talk about an EMF, but not a potential function. We are right bck to a non-conservative E-field and the line integral around a closed loop not being zero.

    The basic problem is that high school mathematics does not permit an adequate description of the physics.

    The only way that you are going to understand this is via a university-level treatment of electrodynamics.
    Okay, thank you for your help.
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