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Induced voltage from a rod sliding on rods in a B-Field

  1. Aug 11, 2015 #1
    1. The problem statement, all variables and given/known data
    Two long rods with distance between them a=100mm, over those two rods there's another rod sliding, friction between them is negligible. The rods are placed in homogenous magnetic field whose B is equal 1T, with direction shown in the picture, between the rods there is voltage generator with constant E, and resistance R=0,1Ω. When mechanical force on rod is zero, then, velocity of rod is v0=10m/s. Represent power of this engine in function of the velocity of the rod and find the mechanical force that acts when power of this engine is strongest, and find that power. Ignore the self-inductance (i believe self-inductance is the correct term).
    Screenshot_1.png

    2. Relevant equations
    e
    ind=∫(v×B)dl

    3. The attempt at a solution

    When there's no mechanical force, then there's only magnetic force due to generator E, but i don't know how to express power using velocity and how can i express it if i don't know is there mechanical force (it says at first that if there's no mechanical force then v is known), i mean is mechanical force important for finding power of the eind "generator" and how can i find the expression wanted?
     
  2. jcsd
  3. Aug 12, 2015 #2

    rude man

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    Well, this one's a bit more involved than your farady disc.
    Questions for you:
    1. what is the net force on the sliding rod, since its velocity is constant?
    So what are the two forces that must cancel each other as the rod slides along?
    Can you determine E from this?
    2. When a force is imposed counter to the direction of v, what is then the net current and force on the sliding rod? Power?
    3. maximize the power the usual way.
     
  4. Aug 12, 2015 #3
    1. As far as i know, rod is moving only when sum of all forces that act on it are not equal zero, since mechanic force is zero, then theres only magnetic force due to generator E, and it's Fmagnetic=IB×a, considering the angles between the two vectors we have Fmagnetic=IBa, I=(E+eind)/R and i have everything but mechanical force and E here but i could find einduced if in case when only magnetic force acts and it's einduced=vBa but that still cant help me find E so i am not really sure what to do with this.

    2. If there's a force opposite to the direction of v, then v could change the direction if the new force is stronger than the one that pushed rod before it. I am not sure for the power, how can i express power? Should it be P=I*(eind+E) because i don't have E yet?

    3. Usual way? By taking derivative?
     
  5. Aug 12, 2015 #4

    rude man

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    That's not what Newton said ....
    In that case the engine would not be providing power, would it? No, direction of v does not change after external mechanical force is applied.
    As I suggested last time, E can be expressed in terms of v (and constants).
    Right
     
  6. Aug 13, 2015 #5
    But then i don't understand. What would happen if there's mechanic force equal by intensity to the magnetic force and with opposite direction?

    Again, i don't understand, why it wouldn't and why it won't change direction, if mechanical force is stronger?

    Well i know that Eind=v×B but that's induced electric field, not E, so i am not sure about this either.
     
  7. Aug 13, 2015 #6

    rude man

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    One thing at a time. You haven't done the first part yet, which is finding E, which you do before a mechanical external force is applied. The rod is moving to the left at 10 m/s with zero current in the sliding rod. Find E.
    You want the motor to provide work against the external force. That means the sliding rod has to move to the left, always. So the external force, applied to the right, cannot be equal to the magnetic force or then v = 0 and no work would be done. You're applying a force somewhere between zero and the mag. force, and the direction of motion is always to the left, as the v arrow shows.
    If you had a motor running clockwise, would you force it to turn counterclockwise with an excessive external force? You'd be doing work on the motor, not the motor doing work on you. Same thing here.
    E is the voltage of the generator, not any electric field. See above.
     
  8. Aug 13, 2015 #7
    Then. i could consider that there are two generators: E and eind so i should have I=(E+eind)/R , but if current is zero, then E=-eind. But i still don't understand how can current be zero. Does this means that every time when i have a generator and only force that acts on rod is magnetic force due to that generator, then eind will always be equal (by intensity) to E?
     
  9. Aug 13, 2015 #8

    rude man

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    I don't know about generalizing the way you did.
    For this problem, the total force, and current, have to be zero, since the sliding rod is not accelerating.
    This is of course before a mechanical force is applied towards the right. At that point, the current has to be finite to counter the mech. force. There will still be no acceleration so the net force on the rod will always be zero.


    .
     
  10. Aug 14, 2015 #9
    Ok, i think i understand now, basically, Fmagnetic=Ia×B=0 , a is not zero, B is not zero, so current must be zero, now, I=(E+eind)/R =0 SO E + eind=0, E=-eind, eind=∫(v×B)dl =vBa=1V which means that E=-1V. Now for the second part. When mechanical force applies, magnetic force won't be zero anymore, that would mean that eind won't be equal by intensity to the E, how can i then express power of engine in function of velocity.
     
  11. Aug 14, 2015 #10

    rude man

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    You're on the right track now! Except for the signs. Set up a coord. system: v = -v i, B = b k and a = -a j so eind=∫(v×B)dl = - vBa and E = + vBa. As shown on your diagram.
    OK: the sliding rod is moving to the left against a force Fmech directed to the right; how much energy does the rod have to produce to move the rod a distance x? and what is the corresponding power?
     
  12. Aug 15, 2015 #11
    I don't know about the energy, but i know that P=Fmag*v (at least i think it should be magnetic force, since the rod has the same direction as magnetic force). Now, since Fmag=IaB, then Fmag=(E+eind)aBv/R. Is this correct?
     
  13. Aug 15, 2015 #12

    rude man

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    Right.
    Almost. Redo your last expression.
    I again urge making units checking a habit. Fmag has units of force which is F whereas (E+eind)aBv/R has units of power which is FLT-1. So your last equation contains an error. Maybe a typo. Really small but still all-important.

    After you fixed that, how about eliminating E and eind with expressions in v?
     
  14. Aug 16, 2015 #13
    I think i know, its:

    P=Fmagv

    now Fmag=IaB

    and I=(E+eind)/R

    so Fmag=aB(E+eind)/R

    and P=aBv(E+eind)/R

    or of i express eind with v (i believe you are asking me to do that since v, a and B are given values) then i have:

    P=aBv(E+avB)/R

    Now, what is the expression for mechanical force, since i have to know it's maximum (i mean it can't be the same as for magnetic, since they are not equal and only way i can express mechanic force is Fmeh=IaB but i doubt it is correct since it's the same as expression for magnetic force)?
     
  15. Aug 16, 2015 #14

    rude man

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    v isw not a "given value" now. v was given only before the mechanical force was applied. Once Fmech is applied, v will be smaller. But your formula for P is now correct if you watch your signs of E and avB.
    Oh but they ARE equal! The rod is still not accelerating, so Fmech + Fmag = 0. So don't be a Doubting Thomas; F = iaB is correct. But watch the signs.
    And you still haven't replaced E with a function of v. (Be careful, v here should really be labeled v0 since it's = 10 m/s whereas v in your P expression is v AFTER the mech. force is applied).
     
  16. Aug 17, 2015 #15
    I know what you mean, because it should be E-eind in the parentheses before i "switched" eind with avB (picture confused me, it shows wrong direction of the eind).

    E=avB since eind=-avB and E=-eind, now i should find derivative of Fmech=-iaB but none of these are variables, how can i do that?
     
  17. Aug 17, 2015 #16

    rude man

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    As I said last time, you need to distinguish between v before the force, and v after. Let us forthwith define v0 as the initial velocity, which is a constant = 10 m/s. Then, v after the force is applied IS a variable; it varies with the mechanical force magnitude.
     
  18. Aug 17, 2015 #17
    Well i guess i could, Since Fmech=-iaB, express B in terms of v like this E =vBa, then B=E/va so Fmech=-iaE/va=-iE/a and then find derivative of this, is this correct?
     
  19. Aug 17, 2015 #18

    rude man

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    You're still not separating v from v0.
     
  20. Aug 18, 2015 #19
    But how can i do that?
     
  21. Aug 18, 2015 #20

    rude man

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    You found the relationship between E and v0 back in post 9! That relationship never changes, even after Fmech is applied. Whereas v is a function of Fmech.
     
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