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Problem with induction, Lorentz force and battery power

  1. Oct 2, 2016 #1
    1. The problem statement, all variables and given/known data

    Hi everybody! I am struggling with an old exam problem, could someone maybe help me out to figure it out? Here is how it goes:

    A rod with resistance ##R = 0.1 \Omega## lays over two parallel tracks (resistance ##\approx 0 \Omega##, ##l=10##cm). A battery is connected between the tracks with a voltage ##U_0 = 15V##. The rod is attached with a massless string to a mass ##m=1##kg hanging down (see picture, the wheel is massless). A current ##I## goes through the closed circuit. A homogeneous magnetic field ##B_0## goes perpendicularly between the tracks. The acceleration due to gravity is to be taken as ##g =10##m/s##^2##.
    1. Give an equation for the current ##I##.
    2. Calculate the velocity at which the rod moves stationary.
    3. Which fraction of the power delivered by the battery is transferred to mechanical work?

    Questions have been translated from German, let me know if something is unclear.

    2. Relevant equations

    Magnetic flux: ##\phi_B = \int \vec{B} d\vec{A}##
    Electromotive force: ##\epsilon = - \frac{d\phi_B}{dt}##
    Lorentz force: ##\vec{F} = I \cdot \vec{L} \times \vec{B}##
    Ohm's law: ##U = R \cdot I##
    Power: ##P = \frac{U^2}{R} = \frac{dW}{dt}##
    Work: ##W = F \cdot s##

    3. The attempt at a solution

    1. I will call the current induced by the battery ##I_0## and the current induced by the magnetic field ##I_{ind}##. First I determine the magnetic flux as a function of time:

    ##\phi_B = \int \vec{B} d\vec{A} = \vec{B} \vec{A}(t) = B_0 (A - \frac{1}{2} g \cdot l \cdot t^2)##
    ##\implies ## electromotive force: ##U_{\epsilon} = - \frac{d\phi}{dt} = -B_0 g l t##
    With Ohm's law I get:
    ##I_{ind} = - \frac{B_0 g l t}{R}##
    which leads me to:
    ##I = |I_{ind}| + |I_0| = \frac{U_0 + B_0 g l t}{R}##

    Now that's neat. It was fast and pretty much straightforward. But I have a problem with this answer: say at time ##t=0## the rod is not moving. Then the acceleration of the rod and the change in area would instantly be ##\frac{1}{2} g t^2##. But what about the Lorentz force? Isn't it acting on the rod in the opposite direction and therefore we would get another equation for the current? A much more complicated one actually. I've given it a try and was unable to get anywhere since everything becomes interdependent.

    Am I overcomplicating the problem with this question? I just feel like this is all wrong if we neglect the Lorentz force...

    2. Assuming my answer in a was correct, I now introduce the Lorentz force acting in the opposite direction of the gravity force. For the velocity to be constant, there must be no net force acting on the rod, that is:

    ##F_L = F_G \iff I \cdot \vec{l} \times \vec{B} = - m \cdot \vec{g} \iff I \cdot l \cdot B_0 = m \cdot g##
    ##\iff \frac{U_0 + B_0 g l t}{R} l \cdot B_0 = m \cdot g##
    ##\implies t = \frac{mgR - U_0 l B_0}{B_0^2 l^2 g}## (time elapsed until velocity is constant)

    Now I must know the acceleration at that time:
    ##\sum F = F_G - F_L = m \cdot a##
    ##\implies a = g - \frac{U_0 + B_0 g l t}{m R} l B_0##

    And the velocity turns out to be:

    ##v = a \cdot t = \Bigg(g - \frac{U_0 + B_0 g l t}{m R} l B_0 \Bigg) \Bigg( \frac{mgR - U_0 l B_0}{B_0^2 l^2 g} \Bigg)##

    What do you guys think? It simplifies a little but not much really, so I prefer to leave it like that for now so that we can discuss the most important steps without having a calculation mistake ruin it all. :biggrin:

    3. This question is tricky, not really sure how to interpret it. Nevertheless I came up with that answer:

    The force component related to the battery is ##F_{U_0} = \frac{U_0 l B_0}{R}##.

    The power delivered by the battery is ##P_{U_0} = U_0 \cdot I_0 = \frac{U_0^2}{R}##.

    Since power is work/time, I guess I can express the power used in mechanical work as:

    ##P_{U_0, mech} = \frac{W_{U_0, mech}}{t} = \frac{F_{U_0} \cdot s}{t} = \frac{U_0 l B_0}{R} \cdot s##

    Then I can take the ratio between the two powers, which leads me to:

    ##\frac{P_{U_0, mech}}{P_{U_0}} = \frac{U_0 l B_0}{R} \cdot s \frac{R}{U_0^2} = \frac{l \cdot B_0}{U_0} s##.

    That's not bad, but what to do with the ##s## now? I believe I cannot insert ##\frac{1}{2} a t^2## in there since the acceleration is not constant. I could put ##v \cdot t## using my results from b, but that doesn't answer the question for ##t < t_{stationary}##... Any suggestion?


    Thanks a lot in advance for your answers, I appreciate. Looking forward to read your suggestions.


    Julien.
     

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  3. Oct 2, 2016 #2

    kuruman

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    You are assuming that the rod is moving at constant acceleration. Is it? Specifically, what is dA/dt? Hint: A = l x therefore dA/dt = ?
     
  4. Oct 2, 2016 #3

    gneill

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    Hi Julien,

    May I ask what level course this question came from? The reason I ask is that if all the interactions taking place in such a setup are to be accounted for then the problem rapidly becomes very complicated indeed. Consider, for example that the current in the rails will also produce a magnetic field that will impart a force on the rod, and it won't be a uniform field.

    It seems odd that no mass has been ascribed to the rod, and that no value for ##B_o## has been given yet a particular value for g is specified. The latter indicates that they expect numerical results, yet that can't be done without the rod mass or magnetic field strength, even assuming a simplified model.
     
  5. Oct 2, 2016 #4
    Hi @kuruman and @gneill and thank you for your answers.

    @kuruman : that's exactly what I am pointing out at the end of my answer for 1, and I am not sure what the answer is. If I consider the Lorentz force when calculating the acceleration, it becomes very complex. I have a hard time believing I should do that, I am supposed to solve such problems in 20 minutes during the exam! If I misunderstood your suggestion, please let me know.

    @gneill I am preparing an exam of electromagnetism, that is 2nd semester of a bachelor of physics. As I said above, I have around 20 mins to solve each problem during the exam. This question comes from an old exam, same course and same teacher as me. I also noticed that nothing is said about the mass of the rod, which I assumed to massless. For ##B_0## that's my mistake: there is actually a convenient value ##B_0 = 1T## given.
    I kind of think my answers for 1 and 2 are what the teacher expects, but then I have this problem with ##s## in question 3.

    Another question I have is concerning question 2. I don't quite get how the rod would keep a stationary velocity after a certain time ##t##. Wouldn't the magnetic flux keep on changing and the acceleration change? Maybe I am wrong, but I would assume that the forces don't remain equal after ##t##.

    Ah also: I don't like how this problem is presented. Unfortunately I must deal with it as it is. :rolleyes:


    Thanks again for your answers.


    Julien.
     
  6. Oct 2, 2016 #5

    kuruman

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    Does this mean that the rod is moving at constant velocity or does it mean that the rod is at rest? In either case, the mass of the rod does not matter because the acceleration is zero and all one needs to do is balance the forces acting on it. We would need the mass of the rod if we were looking for, say, the velocity of the rod as a function of time.

    Also, you did not answer my question, Julien. What is dA/dt? You need it because dΦM/dt= d/dt(A B) and that's related to the induced current.
     
  7. Oct 2, 2016 #6
    @kuruman: to me the question means that the rod is moving at constant velocity, hence ##a = 0##. To answer your question about the surface, I assumed that ##\frac{dA}{dt} = -\frac{1}{2} g t^2##. If instead I take the velocity as ##a = g - \frac{U_0 + B_0 g l t}{mR}l B_0## (##F_G - F_L##), then I guess the answer to your question would be:

    ##\frac{dA}{dt} = - \frac{U_0 + B_0 g l}{mR}l B_0##.

    Is that what you are suggesting?

    I also realised that my answer for 2 is not quite plausible, because I take ##v = a \cdot t## but ##a## is supposed to be equal zero at the given time. That's a problem. :)

    I'll give it some more thought. Let me know if the answer I gave is what you expected.


    Thank you for your answer, I appreciate your help.


    Julien.


    EDIT: Oops I have taken the acceleration I calculated in question 2, but that ain't correct if ##I## is not correct. Here is my problem I guess: ##I## depends on the acceleration which depends on ##I##!
     
    Last edited: Oct 2, 2016
  8. Oct 2, 2016 #7

    gneill

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    Why not start by assuming that the rod is stationary at time t=0 and checking the actual forces at play that instant?
     
  9. Oct 2, 2016 #8

    kuruman

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    That is an incorrect assumption. You are assuming that as soon as the battery is connected at t = 0, the rod accelerates instantaneously to a constant acceleration and then ... the acceleration somehow vanishes when the rod reaches terminal velocity? The answer to my question is much simpler than you think, so I will give some of it away at this point. ## \frac{dA}{dt}=\frac{d}{dt}(lx)=l \frac{dx}{dt} = ?? ## Do you see why this is useful? Also, gneill's post#7 is a useful suggestion to orient you in this problem.
     
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