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Monstrous electromagnetism question.

  1. Jun 2, 2008 #1
    [SOLVED] Monstrous electromagnetism question.

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

    A slope variable: thèta
    Amount of coils: 10.
    Diameter D
    Length L
    Massa M
    Magnetic field B.

    The cilinder has 10 coils wrapped around it, in rectangular form (over the top and bottom of the cilinder). It is placed on a slope with slope thèta, and the coils are horizontal at first. The current goes in a counter-clockwise motion when looked upon from above. There is a magnetic field with strength B, which is perpendicular and upwards. We release the cilinder to all forces. Give the formula for the minimal amound of amps going through the coils to make sure it doesnt keep rolling downwards.

    2. Relevant equations

    rot E = - dB/dt

    3. The attempt at a solution

    Many attempts, many failed. I just had this question on an exam, and man this was a brain teaser. The flux varies sinoïdal due to the rotation of the cilinder. Hence causing an electromotoric force. On the other side, we must find a minimal current to go through the coils, so that when the force pair of the lorentzforce on the coils is strong enough to fight the gravitational force of rolling downwards. This however, can't be at the first moment, because we place the coils horizontally, hence meaning that the force pair cancels eachother out and has no tangential component. Once it begins rotating, it gathers momentum, so we must find a current, so that it cancels out the gravitational force, and it has enough time to reduce the angular speed to zero and reverse it. We also know that the law of faraday states that the amps are influence by the fluctuating magnetic flux through the coils. Hence creating a fluctuating voltage. This will however turn around in signum after rolling 90 degrees, hence it is (IM NOT SURE ABOUT THIS) that the reversal must happen in the first quadrant.

    Any idea's on how to solve this monstrous question?
  2. jcsd
  3. Jun 2, 2008 #2
    Actually, the magnetic field is said to be vertical, which is perpendicular to the horizontal coil. Since the cylinder is still, not rolling, the flux is not sinusoidal. It is constant.
  4. Jun 2, 2008 #3
    Yep, thats what I meant. Once it starts rolling, it starts changing.
    Last edited: Jun 2, 2008
  5. Jun 2, 2008 #4
    Lucky for you, it doesn't roll because we pump some number of amps through the coil to "make sure it doesnt [even start] rolling downwards."

    See, it's really not a monster of a problem, just one force calculation: the force on a conducting coil in a magnetic field. You have the equations for that in Relevant equations, right?
  6. Jun 2, 2008 #5
    Nono. It can start rolling, it just can't fall off.

    So it can start rolling slightly, but then rollback, and maybe even oscillate as far as I know. The clue is to find the MINIMUM amount of amps so it doesn't go uncontrollably downwards.

    It is even impossible not to roll, since the forces by the magnetic field are horizontal, in both directions. This means that it has no torque on the cilinder, and so it HAS to roll. But it just has to brake.

    It's a ***** aint it? :D
  7. Jun 3, 2008 #6
    Wow, I'm sorry for leading you in the wrong direction.
    It does indeed seem like a [female dog] of a problem!
  8. Jun 3, 2008 #7
    Hehe no problem. Anybody else got any ideas?
  9. Jun 4, 2008 #8
    Ok, i had a small idea here, tell me if its mathematically correct:

    If the cilinder starts rolling, lets say with at angle theta, lets let it roll over a certain height difference...

    1/2*mv² + 1/2*I*omega = mgh

    mv² + mr²omega² = 2mgh

    v² + r²omega² = 2gh

    2*r²omega² = 2gh

    (Last step: omega = v / r)

    Now lets look at the height difference h. We know that:

    sin theta = h/s

    Whereas s is the amount of distance covered on the slope.

    omega = root(gh/r²)

    omega = root([g*s*sin(theta)]/r²)

    omega = root(s) * root ([g*sintheta]/r²)

    Now "s" is equal to the amount of distance done on slope. So s=vt:

    omega = root(vt) * root ([g*sintheta]/r²)

    And v= omega-actual*r:

    In which omega-actual is the actual speed in summation of all domega/dt.

    omega = root(omega-actual) * root([g*sintheta]/r)

    Now, I know the torque of the coils are:

    tau = IL2rB*sin(omega-actual*t)

    tau = m*r²*d(omega-coil)/dt


    d(omega-coil)/dt = (IL2B/rm)*sin(omega-actual*t)

    Now i know (omega is the omega of gravity remember!):

    d(omega)/dt = root([g*sintheta]/r) * (1/root(omega-actual) * d(omega-actual)/dt

    Now i know, that the omega-actual is:

    omega-actual = omega0 (=0) + t * d(omega)/dt + t * d(omega-coil)/dt

    Now i know, that omega-actual must be nil at the period/4. (90 degrees) Otherwise we'll get in trouble and d(omega-coil)/dt will start reducing. So:

    0 = P/4 * d(omega)/dt + P/4 * d(omega-coil)/dt

    Now i can scrap the P/4 coefficients.

    d(omega)/dt = - d(omega-coil)/dt

    At t= P/4.

    root([g*sintheta]/r) * (1/root(omega-actual) * d(omega-actual)/dt = - (IL2B/rm)*sin(omega-actual*t)

    the sinus = 1.

    root([g*sintheta]/r) * (1/root(omega-actual) * d(omega-actual)/dt = - (IL2B/rm)

    But now im just getting confused by the /0. I mustve made a mistake somewhere....
  10. Jun 4, 2008 #9


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    Homework Helper

    Hi Swatje,

    I think they want you to use conservation of energy here. The amount of gravitational potential energy decrease is equal to the loop's magnetic potential energy increase. By considering the fact that the potential energy is increasing, you can find the proper orientation of the loop and the direction of the current at both the initial and final points (which are both places where the kinetic energy is zero).
  11. Jun 4, 2008 #10

    Now the minimal amount of I, would be for omega*t to be as large as possible right? So lets say P = period/4, so after 90 degrees:

    mgh = - M . B

    mg*sin(theta)*s = - I * 2*r*L * B * cos (omega*t)

    I = - (m*g*sin(theta)*s)/(2*r*L*B*cos(omega*t))

    Now we want I as small as possible:

    variables are s and omega*t. It also has to happen in first quadrant, and stopping anywhere before p/4, is just extra energy... So i assume:

    s = pi*(1/2)*r

    But then cosine wont work...

    And if i let the cosine be as big as possible, i'll reach zero wt = zero, which is impossible too.

    And does this account for faraday's law? If i run this minimal current through, it'll enlarge due to fluxvariations, and it'll top over the 90degree angle...
    Last edited: Jun 4, 2008
  12. Jun 4, 2008 #11
    Is it even possible?
    Look at the initial configuration: The cylinder is sitting on top of the plane. The net force due to the magnetic field is zero (it cancels along the sides - in fact by symmetry doesn't this always happen?). So at this point the only net force is gravity, so it rolls down the plane a little bit, converting potential energy into kinetic, and probably electromotive energy.

    Now the things to note is that, in the problem:
    There is no excess charge anywhere - all the charge is bounded together in the wire.
    To lift the cylinder would require work to be done on the cylinder.

    So, since the charge is all in the wire it can't be used to push the cylinder uphill. So we'd require the magnetic force to push it. But the magnetic force does no work!
    Hence irrespective of the amount of current flowing through the wire, we couldn't stop it rolling downhill. (Though we would generate a nice EMF, as well as kinetic energy)

    Summary: Since the magnetic field does no work it is impossible.
  13. Jun 4, 2008 #12


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    You'll need to have the energy equation modified a bit to take into account the intial and final magnetic potential energies of the loop:

    (m g h_f - m g h_i) + (- u B \cos (\theta_f) - (- u B \cos(\theta_i) ) = 0

    [itex]\theta_f[/itex] is 90 degrees, but [itex]\theta_i[/itex] is not.

    I think Faraday's law will predicts no induced current at the final point because the rotation of the cylinder is zero there (and so the field, loop area, and relative orientation are not changing).
  14. Jun 4, 2008 #13
    Yes, at time = 0 the netforces cancel eachother out, but when it starts rolling, the forces wont be strictly radial anymore and there will be a tangential component which exerts torque.

    I understand that magnetic force does no work, but can't it oscillate slightly over an angle smaller than 90° degrees? Or will this oscillation start going out of control and eventually tip over the 90 degrees angle?

    Hmm, i see what your getting at.


    mg (hi - hf) - I * L * 2r * B * (cosa theta f - cos theta i)

    theta i = 0


    mg (hi - hf) - I * L * 2r * B * cos theta f = 0

    Which is pretty much the same equation i had just before, where my h = hi - hf.

    And faraday doesnt predict any current at final point indeed, does that also mean it doesnt influence the minimal initial current???
    Last edited: Jun 4, 2008
  15. Jun 4, 2008 #14


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    Hi fantispug,

    A magnetic field does no work; but when a magnetic field affects a magnetic moment there is work being done. That's the point of having a magnetic potential energy. In this case I believe the battery which is driving the current is the source of the work.
  16. Jun 4, 2008 #15


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    Since theta_i=0, and theta_f=90, you get:

    mg (hf - hi) - ( 0 - I L 2r B) =0

    m g (delta h) + I L 2r B = 0

    where (delta h) is negative.
  17. Jun 4, 2008 #16
    Oh yes i see, small mistake ;).

    so then:

    I = - (m g delta h ) / (L * 2r * B)

    and delta h, will approxiamtely be:

    - sin theta * PI * r * 1/2


    I = (m*g*pi) / ( 4 * L * B)

    And you can discard any increase by faraday, because it will eventually have to stop rolling anyways, and then I = I original.
    Last edited: Jun 4, 2008
  18. Jun 4, 2008 #17


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    Minor point: I think you dropped the sin theta (that came from the change in height) from your final result.
  19. Jun 4, 2008 #18
    Yep thx, and I also forget to add a 10 of the amount of coils.
  20. Jun 4, 2008 #19
    Ah, right, I was wondering about magnetic potential energy. The battery being a source of the work probably makes sense. I'm still having trouble visualising the situation though.... if the current is reasonable I might try to construct it.
  21. Jun 4, 2008 #20
    Im afraid your gonna need alot of tesla's to construct it. 1 maybe 1.5 tesla.
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