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Electromagnetic force and SHM

  1. Oct 24, 2015 #1
    1. The problem statement, all variables and given/known data

    We take a horizontal copper bar with length of 20cm and attach it from the middle to a vertical spring which mass is neglected and has a spring constant K, we apply a horizontal magnetic field with magnitude 1/2 T and have a 10A current run in the copper bar. The bar rests after the spring is stretched by 10cm of its original length before turning on the current.

    1 Define the forces affecting the system at rest knowing that the electromagnetic force is downards
    2. Find K.

    2. Relevant equations



    3. The attempt at a solution

    I honestly have no idea what to do, the wording is somewhat confusing, here's how I ran the problem in my mind:

    When we attach the bar to the spring, it stretches by length X0=m*g/k, and the system is at rest. I don't know what x0 is. The forces affecting the system then are Fs0 and Fs0' which cancel out, plus W (weight) and R (normal force) which also cancel out. Is this correct?

    When we turn on the current, the electromagnetic force keeps pulling the body down, and the restoring force keeps trying to bring it up, thus, they also cancel out along the other forces. The forces affecting the system now are F(electromagnetic), F(Restoring force), W (Weight), R(Normal force), Fs0 and Fs'0. Is this correct?


    So I have 2 rest points, one before the current, and one after the current? How am I supposed to find K?

    If I say that F(electromagnetic) = F (Restoring force) and then plugin the values:

    I*L*B*= -*K*X

    10*2*10^-1*1/2=-K*X

    1 = -K*X

    K=-1/X

    K=-1/-10^-1 = 10 N.m^-1

    This doesn't really sound correct to me, is it?
     
  2. jcsd
  3. Oct 24, 2015 #2

    ehild

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    What do you think the relevant equations are concerning the problem?

    What are the forces Fs0 and Fs'0, and what is the normal force? A figure would be useful with the forces shown.
    As for the two equilibrium points, yes, there was one before the current has been switched on, and the other one when the current is flowing.
    You do not know what are the equilibrium lengths of the spring, but you know the difference of the equilibrium lengths: The bar rests after the spring is stretched by 10cm of its original length before turning on the current. So the spring had some length before turning on the current, and stretches by 10 cm after the current is on - you are on the right way.
     
  4. Oct 24, 2015 #3
    Fs0 is the spring's pull on the bar, and Fs'0 is the bar's pull on the spring, x0 is the initial stretch, x is the stretch after the current is turned on, you can just ignore what I said about the normal force.

    What I did was apply Newton's second law to the bar only after the current was turned on:

    The forces acting on it are: Weight, Electromagnetic force, and Fs0 (the spring's pull on the bar).

    W + F + Fs0 = 0

    Axis is directed downwards:

    m*g - k(x0+x) + F=0

    k*x0 - k(x0+x) + F =0

    k (x0 -x0 - x) + F = 0

    K*X = F

    I already know F, and X = 10cm, so I calculated K. This doesn't sound correct, but it's all I've got.
     
  5. Oct 24, 2015 #4

    ehild

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    Think: The magnetic force is very small, only one N. The weight of a 0.1 kg body. And it stretches the spring by 10 cm, which is quite much. So it is a weak spring.
     
  6. Oct 24, 2015 #5
    So the answer's correct?

    Never really thought about checking if the values make sense.
     
  7. Oct 24, 2015 #6

    ehild

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    It looks correct.
     
  8. Oct 25, 2015 #7
    Okay, I asked my teacher and he said you don't consider X0, you only take the displacement caused by the electromagnetic force, not the weight. His answer was k = 30

    I don't really get it, can someone explain what I did wrong and what I should have done?
     
  9. Oct 25, 2015 #8

    gneill

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    A one Newton force stretched the spring by 10 cm. How did he arrive at a k of 30 N/m o_O ?? Methinks something's gone awry.
     
  10. Oct 25, 2015 #9

    ehild

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    You did not use x0 either. I do not know how you teacher got that value for k. Did you copied the problem correctly? Was not the length of the bar 60 cm?
     
  11. Oct 25, 2015 #10
    No idea, I only talked briefly to him, I'll ask him again tomorrow and he should explain it fully. (Other students also got 30 somehow).

    Yes, I copied it correctly.
     
  12. Oct 25, 2015 #11
    Okay, nevermind, I figured out how they reached 30, although my answer seems more correct to me than their, I'm sure people here can tell which answer is correct.

    What the teacher did: (M is 1/5 KG)

    Forces affecting the bar:

    W + Fs0 + F(electromagnetic) =0

    Axis is directed downwards

    W - Fs0 + F(electromagnetic) =0

    m*g - k*x + F(electromagnetic)=0

    1/5 * 10 - k*x + F(electromagnetic) =0

    (he only substitutes x = 10 cm, neglecting the x0 before turning on the current)

    2 - k*10^-1 + F(electromagnetic) =0

    K = F(electromagnetic)+2/10^-1

    F(em) is 1 as calculated before

    K = 3/10^-1 = 30 N.m^-1


    So, which method is correct?
     
  13. Oct 25, 2015 #12

    gneill

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    Where did the 1/5 kg mass come from? It wasn't in the problem statement. And if the spring constant turns out to be 30 N/m, then the initial displacement (hanging mass, no magnetic force) should have been about 6.5 cm, not the 10 cm that was given. Right there the solution looks implausible.
     
  14. Oct 25, 2015 #13

    ehild

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    You and the teacher solved different problems.
    The teacher's version could have been
    "We take a horizontal copper bar of 1/5 kg with length of 20cm and attach it from the middle to a vertical spring which mass is neglected and has a spring constant K, We apply a horizontal magnetic field with magnitude 1/2 T and have a 10A current run in the copper bar. The bar rests after the spring is stretched by 10cm of its original length. before turning on the current."
    So he did not neglect the original change of length before the current was turned on, but included it into the 10 cm.
    Now it is your turn to look after, where did that 1/5 kg mass come from.
    The last sentence in your OP is ambiguous. The spring was stretched before turning on the current. So "The bar rests after the spring is stretched by 10cm of its original length before turning on the current." can be understood also that the spring was stretched by 10 cm because of its weight , but then no need to give the magnetic field and current. Or it can be understood that the 10 cm came because of the electromagnetic force. But it is impossible to understand it that the 10 cm stretching was caused by both forces - gravity and electromagnetic.
     
  15. Oct 25, 2015 #14
    It's a multi task problem with several questions, the first ones are all about electromagnetic stuff and no mechanics involved, I copied the mechanics part but forgot to add the mass (didn't think it's relevant).

    Yes, the wording is indeed confusing, the way I understood the last sentence is that the 10cm stretch happens due to the electromagnetic force, and a previous, unknown stretch happens because of weight.
     
  16. Oct 25, 2015 #15

    ehild

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    Always copy the whole problem as it is written. I even asked you if you copied the problem correctly. Do not leave out any data.
     
  17. Oct 26, 2015 #16
    Teacher considered the 10cm displacement to be caused by both weight and EM force and said to scrap the "before current is turned on" part. But even then I'd still solve it the same way I did first, whatever.
     
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