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Voltage across an inductor

  1. Nov 1, 2009 #1
    Reading a different homework problem here made me realize I don't understand somethings about inductors. I made a simple problem that focuses just on the inductor to help me learn. Please do help me solve this problem.

    1. The problem statement, all variables and given/known data

    Consider this:
    15i7zab.gif
    The loop has inductance L, no resistance, and that initially there is no current in the loop. Then at time t=0, a current source is used to increase the current at a steady rate (I = a t).

    Current goes up the green wire, around the blue loop, and then up the red wire. Additionally, consider a voltmeter connected with the ground on the red point, and the "positive" terminal on the green point.

    What does the volt meter read?

    2. Relevant equations

    I'm not sure what you want here.

    I believe a voltmeter reads (with leads on points 'a' and 'b'):
    [tex] Voltage = - \int_a^b \mathbf{E} \cdot d\mathbf{l} [/tex]


    All of electrodynamics is in these five equations:
    [tex] \nabla \cdot \mathbf{E} = \frac {\rho} {\varepsilon_0}[/tex]
    [tex] \nabla \times \mathbf{E} = -\frac{\partial \mathbf{B}} {\partial t} [/tex]
    [tex] \nabla \cdot \mathbf{B} = 0 [/tex]
    [tex] \nabla \times \mathbf{B} = \mu_0\mathbf{J} + \mu_0 \varepsilon_0 \frac{\partial \mathbf{E}} {\partial t} [/tex]
    [tex] \mathbf{F} = q (\mathbf{E} + \mathbf{v}\times\mathbf{B})[/tex]

    The definition of flux, and inductance:
    [tex] \Phi = \int_S \mathbf{B} \cdot d\mathbf{a}[/tex]
    [tex] \Phi = L I [/tex]

    In case it is useful, there is also the definition of the potentials:
    [tex] \mathbf{E} = - \nabla V - \frac{\partial}{\partial t} \mathbf{A}[/tex]
    [tex] \mathbf{B} = \nabla \times \mathbf{A}[/tex]
    We will use Coloumb gauge here:
    [tex] \nabla \cdot \mathbf{A} = 0[/tex]


    3. The attempt at a solution

    There are a few steps in my logic, please let me know if they are correct.

    part 1]
    Using Maxwell's equations, when the current starts increasing a magnetic field will be generated in the upward direction inside the loop in that diagram. In other words, the magnetic flux of the loop will increase with time (with 'up' defined as positive).
    Correct?

    part 2]
    Due to the negative sign in Del.E = - (d/dt) B, an electric field will be generated to oppose this change in flux (this is also known as Lenz's law). So the electric field will be pointing against the current around the loop. Correct?

    part 3]
    Similarly, this means the integral of - E.dl from the red to the green point is negative.
    This seems to argue the voltmeter would read negative.
    Correct?

    If I actually plug in all the math along these steps, I get:
    Voltage = - L dI/dt = - L a


    However, I know that this is wrong, as the answer I should be getting is:
    Voltage = L dI/dt = L a
    and furthermore, if it actually was negative, then the current would be flowing against the electric field which doesn't make any sense.

    What the heck is going on here?
    Which part of logic above is incorrect?
     

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  3. Nov 1, 2009 #2

    Andrew Mason

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    To begin, try analysing the problem qualitatively.

    If the wire, including the loop, has no resistance, there can be no voltage difference if there is no change in current. So the voltage is entirely due to the induced voltage in the circuit. This is a function of the rate of change of the magnetic flux enclosed by the coil (Faraday's law). The magnetic flux enclosed by the coil is a function of the current in the coil (Ampere's law). You can use Lenz' law to determine the direction of the induced voltage.


    Now, to do the quantitative analysis.

    Apply Ampere's law to the single coil:

    [tex]\oint B\cdot ds = \mu_0 I_{loop}[/tex]

    By symmetry, the LS is: [itex]2\pi r B[/itex]. I is time dependent. So applying Ampere's law you can get the expression for B in terms of time.

    You can now determine dB/dt and apply Faraday's law to determine the induced voltage. Use:

    [tex]Emf_{induced} = -d\phi/dt = -AdB/dt[/tex]

    AM
     
  4. Nov 1, 2009 #3
    I did provide qualitative analysis.
    Since I get the correct magnitude with detailed analysis, but the sign of the answer is wrong, it looks like I'm making a sign convention error of some kind in my qualitative analysis.

    Can you look over my three qualitative steps and tell me if you agree with the conclusions of each one?
    In particular, the step that determines the direction of the electric field is bothering me. You seem to get the same incorrect answer with your logic as well. I don't understand what is wrong here

    Do you agree that the induced electric field is going clockwise around the loop?
     
  5. Nov 1, 2009 #4

    Andrew Mason

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    Yes.

    The induced voltage has to be opposite to current. The only thing limiting the current is the induced emf which has to be in the opposite direction to the current. There is no other voltage contribution.

    The current flows from green to red. This is the direction that positive charges move. Since the induced voltage is opposite to this, the green terminal must be negative and the red terminal positive, so it will show a negative voltage.

    The direction of the voltage depends on the sign of a. If a>0, which is stated to be the case here, the induced emf is opposite to the increasing current. if a<0, which is not the case here, the induced emf is in the same direction as the decreasing current. This is the effect of Emf = -Ldi/dt. If di/dt>0, the induced emf is opposite to the current. If di/dt<0 then the emf is in the same direction as the current.

    AM
     
    Last edited: Nov 1, 2009
  6. Nov 2, 2009 #5
    Then you are making a mistake as well.
    For that is incorrect.

    Imagine a resistor and inductor in series across a voltage source. If V(t) = 0 for t<0, and V(t) = V_0 (a positive constant) for t>0, then you are claiming the voltage drop across with inductor will be negative for t>0 and thus more current will flow through the resistor than if the inductor was not there. You are claiming an inductor will AMPLIFY current spikes, when in reality they suppress them.


    Can anyone out there see the mistake that Andrew and I are making?
    What are we forgetting?
     
  7. Nov 2, 2009 #6

    ehild

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    Justin,

    The problem with an inductor is that the electric field is not conservative inside it. The line integral along a closed path ( along a circle just beside the wire loop) is

    [tex]\oint{E_sds}=-\frac{\partial \Phi} {\partial t}[/tex]


    To use the concept of potential and voltage, you have to choose such integration path which exclude the domain with time dependent magnetic field. You want to know the reading of a voltmeter connected to the terminal of your wire loop. Choose a path which goes along the loop (but not the wire itself) and along a resistor that represents the voltmeter, but does not include the magnetic field. The integral of the electric field along this closed curve makes zero.

    [tex]\oint{E_sds}=\int _A^B{E_sds} +\int_B^A{E_sds}= 0
    [/tex]

    As the tangential component of the electric field is continuous, it is the same at both circular path, and also it is the same inside and outside the resistor.

    [tex]\int _A^B{E_sds}=-\frac{\partial \Phi} {\partial t}[/tex]

    for the circular path, and

    [tex]\int_{B'}^{A'}{E_rdl} = \int_{B'}^{A'}{-\nabla{U}dl}=\int_{B'}^{A'}{-dU}=U(B')-U(A') [/tex]

    for the resistor. We can assume zero electric field along the connecting wires, so A and A' are at the same potential and so are B and B'. The sum of the integrals is zero, so

    [tex] U(B)-U(A) =\frac{\partial \Phi} {\partial t}[/tex]

    The potential at B is positive with respect to A in case of increasing flux.


    ehild
     
    Last edited: Jun 29, 2010
  8. Nov 2, 2009 #7
    You seem to be arguing that the voltage a voltmeter reads is:
    [tex] voltage = -\int_A^B \mathbf{E} \cdot d\mathbf{l}[/tex]
    for the "conservative" part of E, and
    [tex] voltage = + \int_A^B \mathbf{E} \cdot d\mathbf{l}[/tex]
    for the "non-conservative" part of E.

    Is that indeed what you are saying?
    I see that does fix the sign problem, but does not seem justified to me. For to reverse the sign in the second one, you demand that a path is chosen such that [tex]\oint \mathbf{E} \cdot d \mathbf{l} = 0[/tex] which doesn't seem physically motivated (or even clear if such a path exists here) when there are 'non-conservative' electric fields.
     
  9. Nov 2, 2009 #8

    ehild

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    No. I say that potential and voltage is defined for a conservative field only. You can calculate line integral, but it can not be connected to potential difference in the domain, where the this integral does depend on the path.

    See my picture. Far away from the loop, the magnetic field vanishes. If you have a coil instead of a single loop, the magnetic field is practically zero beside the coil, and it vanishes soon in the axis of the coil, to. So you can choose closed paths which do not enclose magnetic field and can define voltage in the domain where the magnetic field vanishes.

    ehild

    ehild
     
  10. Nov 2, 2009 #9

    ehild

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    Voltage is defined between points A and B as drop of the potential when walking from A to B.

    The electromotive force of an ideal voltage source is opposite to the voltage across its terminals.

    Look at a simple circuit containing a battery and a resistor. The current flows out of the battery at the positive terminal, goes through the resistor and enters the battery at the negative terminal and it flows from - to + inside the battery. Does it mean that the electric field inside the battery points from - to +? That the "+" of the battery is negative with respect to the "-" terminal? The truth is that we do not know the forces acting on the charged particles inside the battery. There are chemical reactions, diffusion, electric field. We know only the result, that one terminal is + and the other is - and going from - to +, the potential will increase by the emf, so the drop of potential is negative.

    ehild
     
  11. Nov 2, 2009 #10

    Andrew Mason

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    Are you saying that Lenz' law is incorrect?

    No. I am not sure how you conclude this from what I said. I am saying the current will be LESS because of the induced voltage. I said the voltage opposes the increase in current - it tries to reduce it.

    There is a time difference between voltage and current here that you seem to be missing. In a purely inductive circuit, such as this, voltage and current are not in phase.

    AM
     
  12. Nov 2, 2009 #11

    Andrew Mason

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    This may simply be a problem with the convention for positive and negative.

    In the circuit, current is flowing at an increasing rate from the green to red. This is the direction that positive charges will move. The induced voltage opposes the increase in current so it tries to oppose the current flow ie. it tends to make the positive charges flow toward the green side. In other words, it tries to make the negative charges flow toward the red.

    Now the question is what is the direction of the potential that will cause that to occur? The induced E field is in the direction toward the green end, as that is the direction that positive charges are pushed by the induced voltage. Since positive charges move from positive to negative, I said that the potential on the green side would be more negative. I may be wrong on that but that is why I said it would be more negative.

    AM
     
  13. Nov 2, 2009 #12

    jambaugh

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    It may help to look at the circuit causally. If you want to induce a current in the loop flowing from green to red you must apply a voltage more positive on the green end i.e. positive as the meter reads since the red is ground.

    Without self inductance and resistance any voltage will cause infinite current flow...(i.e. seemingly negligible resistance will come into play and you get massive current). The inductance cannot prevent the current from growing toward infinity but it will keep it from happening instantaneously. The fixed voltage across the inductor of fixed inductance causes a linear increase in current as described.

    The applied voltage must match the back voltage of the increasing upward B field which induces the back voltage (opposite the direction of current flow but not opposing the current as such). The inductance opposes a change in the current just as a resistor resists the current itself. The measured voltage is the reaction to the effect on the charges in both cases, not the effect itself. That is to say the voltage is induced externally so as to maintain the increasing current in spite of the inductance (or for a resistor to maintain the current in spite of the resistance).

    Imagine instead now that the same loop is connected externally to a resistor and the same increasing B field effected by an external source. It is in that case that an opposing current (current inducing an opposite B field) would be induced but it would still yield a positive reading on the voltmeter as this current flows from red to green through the loop and then back from green to red externally through my appended resistor.
     
  14. Nov 2, 2009 #13

    Andrew Mason

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    This is not true. Current can flow through the wire without a potential difference from green to red. In fact, this is the case where the current is not changing since there is no resistance. V = IR = 0.

    The rule is simple: the increasing current in the loop creates an induced emf that opposes the increase in current. The emf measured between the green and red ends is opposite in direction to the direction of the current.

    AM
     
  15. Nov 2, 2009 #14

    jambaugh

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    Yes but there will also be a B field. My statement presupposes the default starting condition of no current and no B field. To cause the current and its necessary corresponding B field from a starting position of neither, one must apply a voltage (or as I mentioned in the second example an external source of the B field)
    And yet this simple rule confused the OP. That is because the "opposition" is applied to the external source of the current. The meter measures that voltage which is effecting the given increase in current not opposing it. This same voltage pushing the higher current is also trying to push current back through the external emf source e.g. a battery. It is in this external sense that it "opposes" the current.

    Look at the mechanical analogue. Inductance is analogous to inertia and potential difference to force (charge=>displacement,current=> velocity, resistance=>drag, capacitance=>elasticity).
    The "opposing emf" of which you speak is like the inertial force (in the sense of http://en.wikipedia.org/wiki/D%27Alembert%27s_principle" [Broken]) of an accelerated mass. It is not the applied force accelerating the mass. It is the applied force which is the analogue of the measured voltage causing the change in current.

    This is the source of the confusion as to sign since the two "forces" are by the defining constraints equal and opposite. It is an improper equivocation of the causal order and logical order.
     
    Last edited by a moderator: May 4, 2017
  16. Nov 2, 2009 #15

    Andrew Mason

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    There is no voltage applied here between the green and red points. There is just the induced emf.

    This makes no sense. The induced emf opposes the increase in current. It can only work in one direction. There is no applied voltage here and we cannot impute one.

    AM
     
  17. Nov 2, 2009 #16
    You claimed that the voltage would be negative across the inductor. Therefore, the voltage drop across the resistor would be LARGER than if the inductor was not there ... and therefore there would be MORE current.

    To oppose the current, the voltage would have to be positive.
    Yet if we use maxwell's equations and calculate [tex]- \int \mathbf{E} \cdot d\mathbf{l}[/tex] we get a negative number.

    Also, since there seems to be disagreement between people here on what is even meany by voltage, let me define what I mean by it. Voltage is what a volt meter reads between two points. Note that this is clearly different than the scalar potential difference between two points.


    Ehild,
    Your solution seems to be based on saying that a voltmeter doesn't necessarily read
    [tex]- \int \mathbf{E} \cdot d\mathbf{l}[/tex]
    I am understanding that much at least?
    Unfortunately, I am not understanding what you are saying a voltmeter does read.

    The old style voltmeters were actually just a sensitive ammeter with a large resistor. So I could say the voltmeter reads what current flows through that resistor (and I think this is what you are hinting at). If I calculate E.dl across that resistor, that would be the voltage? But if I take the voltage drop across a 'voltmeter connecting wire' to be zero. then all this does is again give me the E.dl across the inductor which is giving the incorrect sign.


    Furthermore, almost everyone seems to agree the current is going AGAINST the electric field. How is that even possible?
     
  18. Nov 2, 2009 #17

    Andrew Mason

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    Actually, I said that the induced emf would oppose the increase in current. I explained why I said that makes the red positive and the green negative.

    That does mean that there is a potential difference across the inductor and its magnitude is certainly greater than if the inductor was not there. If the inductor was not there, the potential difference would be zero.

    Ok. Just state what your convention is and we can go with that. If current is the direction positive charges move then the field created by the induced voltage must be toward the green end. In the convention I use, that makes the red end positive. I am not sure what your convention is.

    Why would it be different? The voltage is the potential energy per unit charge (in Joules/coulomb) of one point relative to the other.


    The induced emf is trying to stop the increase in current. It is not causing the current to flow. If it were the other way around, you would have a first order perpetual motion machine.

    AM
     
  19. Nov 2, 2009 #18

    jambaugh

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    The problem states "a current source is applied". Now you can supply available charge all day but that does not constitute a current source. A current source must apply sufficient voltage to achieve the desired current through the circuit. Given the correct measured voltage is such that green minus red is positive, given that voltage drives positive charge from higher potential to lower potential, how can you say there is no applied voltage here?

    To effect the situation described in the problem you must either 'impute' an applied voltage or 'impute' an applied varying external B field. Take your pick we can work the causal chain either way. This appears to be a problem of self inductance ("applied current source") and not one of external inductance ("induced current") and so I would think 'imputing' an applied voltage is far from controversial.

    Do you disagree with the assertion that the green points in the diagram are at a higher voltage than the red?
     
  20. Nov 3, 2009 #19

    ehild

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    Justin,

    Before you speak about voltage, learn what it is.

    Voltage is defined between two point in a conservative electric field. Note the term "conservative" and "electric".

    When a point charge moves from point A to B, the work of the electric field is

    [tex]W_A_B=q\int_A^B{\vector{E}\cdot\vctor{dr}}[/tex]

    Only if this work is independent of the path taken - the electric field is conservative - can we speak about electric potential [itex]V(\vector{r})[/itex]. The electric field strength is the negative gradient of this potential function, [itex]\vector{E(\vector{r})}=-\nabla{V(\vector{r})}[/itex]. In this case,

    [tex]W_A_B=-q\int_A^B{\nabla{V(\vector{r})}\cdot\vector{dr}}=-q\int_A^B{dV}=q(V(A)-V(B))[/tex].

    The voltage VAB is defined in conservative electric field between two points A and B as the work of the electric field on a positive unit charge while it moves from A to B. The voltage is negative potential difference.

    VAB=V(A)-V(B).

    The sign of voltage depends on the direction of motion. You can not speak about voltage in general without indicating the direction of motion. VAB means the potential drop from A to B. It can be confusing when people forget about it. It is better to use potential than voltage when applying Kirchhoff's Loop Law in solving circuit problems.

    Potential and voltage is defined in conservative electric fields. Here the line integral of E along a closed path is zero. If E has curl, this is not true, and you can not use the term "voltage" in a circuit which encloses a domain with non-vanishing curl(E). In conservative fields, you can get the force acting on a point charge from the gradient of the potential, you can calculate the magnitude and direction of current from the potential difference, but is is not true when there are other forces acting on the charge.

    In a simple battery, the motion of ions is governed by electrochemical forces. You can not calculate voltage across the terminals as line integral along a path which leads inside the battery. You do not know the electric field inside the battery and you do not need to know about it. All you need to know is emf, the voltage you can measure with a voltmeter across the terminals -outside of the battery.

    Connect a resistor to the battery, current will flow through the wires and through the resistor. Follow the current flow and trace the potential. If you start from the positive pole the potential will drop by IR on the resistor, and rise by the emf from the negative pole to the positive one across the battery. The net change of potential is zero, so emf = IR.
    Notice that the current flows from the negative pole to the positive one inside the battery. Does it mean that the electric field points from negative to positive? Does it mean that the potential drops from the negative pole from the positive one? This would contradict the + and - signs written on the battery!
    The motion of charged particles is beyond electricity inside the battery. You can not calculate voltage by integrating along a path which is inside the battery.

    A coil is like a battery. Assume first that it is not an external current, that causes the flux change in time.
    The time-varying magnetic field is beyond stationary electric fields and currents. Current flows as the time-varying magnetic field causes non-conservative electric field and that drives the electrons round the loops and out of the coil, across a voltmeter or resistor, causing a potential drop . The sign of the potential change is determined by the direction of current: Across the resistor or a voltmeter, the current flows from the place at positive potential towards the negative one.

    Now it is your problem: The magnetic field is governed by an external current. The sign of emf of a voltage source is determined with respect to the current. It is negative if the current flows in at the positive pole. You have learnt what is the emf of a coil: it is negative with respect to the current if the current is increasing. Increasing current causes negative emf: that is, the current flows into the coil at the positive pole.


    ehild
     
    Last edited: Nov 3, 2009
  21. Nov 3, 2009 #20
    You seemed to have missed my point.
    Please reread:
    "Imagine a resistor and inductor in series across a voltage source. If V(t) = 0 for t<0, and V(t) = V_0 (a positive constant) for t>0, then you are claiming the voltage drop across with inductor will be negative for t>0 and thus more current will flow through the resistor than if the inductor was not there. You are claiming an inductor will AMPLIFY current spikes, when in reality they suppress them."

    You cannot claim the voltage drop across the inductor is negative AND that the inductor is suppressing the change in current.

    Why would it be different? Because if a volt meter actually read the difference in scalar potential, then the measurement would violate gauge invariance.

    There must be a force causing the current to flow. The only forces in that wire are electrical. So where is the electric field that is opposing this "induced electric field" and how do I calculate it?


    I defined what I meant by voltage.
    All you seem to mean by voltage is the "scalar potential". But that is defined even when the electric field is not "conservative". Please refer to my first post and let me know if you disagree with the definitions there.

    I will use the term "voltmeter reading" from now on instead of "voltage" to prevent confusion.

    No,
    [tex] \mathbf{E} = -\nabla V - \frac{\partial}{\partial t} \mathbf{A} [/tex]
    The scalar and vector potential always "exist" as they can be defined in any situation.

    Furthermore, the work to move a charge against electromagnetic forces is always Integral E.dl. Yes it becomes path dependent if the fields are not conservative, but that statement about the work is still true.

    Engineers talk about the voltage across inductors all the time.
    Please define what you mean by that term, as I have defined what I meant when I used it, but you clearly are using a different meaning without telling me. (It sounds like you mean "scalar potential", but for the reasons stated above, I am not sure since your statements are still not correct in that context either.)

    True, there are other forces involved in a battery. But the "voltmeter reading" could still be calculated in theory if the fields were known. Since we are dealing with situations where the only forces are eletrical forces, we don't need to worry about such calculational problems.

    No, because there are other forces involved.

    The magnetic field does not drive the electrons. The change in magnetic field is a source for an electric field. Only the electric field can do work on charges.

    Since the current is going against the "induced" electric field, there must be another electric field countering this or something. What is driving the current? It must be an electric field, and it can't be the electric field induced by the changing magnetic field.
     
    Last edited: Nov 3, 2009
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