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How much will this wire shift?(not sure if my solution is right)

  1. Nov 14, 2013 #1
    Cirquit homework

    1. The problem statement, all variables and given/known data
    A wire hangs loosely in a magnetic field, it is free to move up or down, the field is 90° on the wire. So the wire gets a charge of 2C over a period of time that is not stated. find the distance it moves during this time.

    Wire has length 0,25cm, mass m, field is 0,4 T.

    2. Relevant equations

    F=ma, F=b*q*i
    3. The attempt at a solution

    F=ma=b*I*i=b*2C/t*l <-> a=h/t^2=b*2C/t*l <-> h = distance moved = (b*2c*l)*t. All I can get is a function of time

    how to I get a numeric answer that is not a function?

    I have a possible solution, though. Magnetic force is excerted over the given time t so t shortenes out and the charge unit coloumb too I guess? so I'd end up with the number h = 0,4 * 2* 0,25 m= 20 centimeters moved.

    Since all the units make it add up, I am not sure if this is correct though.

    Nevermind I have solved it.
    Last edited: Nov 14, 2013
  2. jcsd
  3. Nov 14, 2013 #2


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    I'm not entirely sure I understand the set-up. If it's hanging, why does it say it can move up or down? Wouldn't it be free to swing left or right? But then, it wouldn't move a 'distance' - it would move through an angle.
    Have you quoted the question exactly as posed?
  4. Nov 14, 2013 #3
    The magnetic field is perpendicular to the wire, hence the force is in only the Y axis, no force is pushing in the X-axis directions. The angle is 90° which means the vector part that pushes it to the sides is 0 or if you want to be specific c(90)*Fb=0. And no I have not quoted it correctly, I have finished the solution but if you are interested I can give you the correct wording of it.

    I'll put this scematic I have drawn for my report, of the system, maybe it'll give you a slightly clearer picture of the concept.

    (I apologize for my horrid handwriting in advance)

    ooops, I confused HORIZONTAL to PERPENDICULAR, I'm not a native enrish speeker.

    Attached Files:

  5. Nov 14, 2013 #4

    rude man

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    Very strange problem.

    When the wire stops moving there is no emf generated from one end to the other; there is never a net charge added to or subtracted from the wire; and the electric field inside and outside the wire are both zero.
  6. Nov 14, 2013 #5


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    OK, thanks for the clarification. But yes, pls post your solution. I'm still as puzzled as rude man.
    If gravity is to be taken into account, then it won't move up at all unless the force is enough to overcome gravity, and that will depend on the current. If gravity is to be ignored then the force leading to the displacement has done work, yet there's no gain in PE, so where has the work gone? Presumably it still has KE:
    a = IBL/m; v = BL∫Idt/m = QBL/m; s = BL∫Q(t)dt/m = ?
  7. Nov 14, 2013 #6
    Okay my solution is the following, and no there is no gravity included. But the mass of the moving parts is 5 kilograms. And we name the time the event happens in, T.

    Give the movable part including the wire mass m. The force on each instant is d/dt(Ft)=d/dt(B*l*I) this is the magnetic force on the wire. Now we know that l, m, B are constants so we get:

    d/dt(Ft)=m*d/dt(a)=B*l*d/dt(I) <-> ma' = B*l*I'. Now, acknowledge the fact that over the time "t" a charge of 2 coulombs passes through the wire, and since we assume the rate of it passing through is a constant, the current on each instant is the same as the current over the whole, that is " I' = 2C / T.(*) Now we get, using (*) that the acceleration of the wire is:

    a' = B*l*(d/dt(2C/t))/m=B*l*2C/T^2*(1/m). Lets acknowledge the fact that since the current is a constant, the acceleration is also a constant so that the distance moved over the given time "T" which is h/T^2 is equal to a'. Now we get:

    h/T^2=b*l*2C/T^2*1/m <-> h=0,4 T * 0,25 m * 2 coulomb* 1/5kg = 0,04 meters.

    This shows that the distance moved when the charge 2C is moved through the wire, is 4 centimeters. Regardless of the time it takes.

    There might be something off with my solution but I'll see when my professor revises it.
    Last edited: Nov 14, 2013
  8. Nov 14, 2013 #7


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    Through all that, I don't know why you show d/dt. Shouldn't it read:
    The force on each instant is F(t)=B*l*I(t). Now we know that l, m, B are constants so we get:
    F(t)=m*a=B*l*I(t) <-> ma = B*l*I.
    I = 2C / T.(*)
    I assume you meant a = h/T2, but that's not quite right. If the acceleration is constant (and we do not know that it is) a = 2h/T2.
    Where did you get the extra 1/T on the right from?
    2h/T2 = B*l*I/m = b*l*(Q/T)/m; h = 2 B l Q T / m
  9. Nov 15, 2013 #8

    d/dt is the force at each instant. I do this because I can then get that extra 1/T you were asking for.

    The magnetic field is a constant. And the current is a constant, too. Thus the acceleration is also a constant.

    The extra 1/T comes from d/dt(2c/T) which equals 2C/T

    I got the right answer, I just recieved the solution from my school, albeit it is quite, simpler then my approach.
  10. Nov 15, 2013 #9


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    No, F(t) = B*l*I(t) is the force at at instant t. dF/dt would give you a rate of change of force. You cannot go inserting extra d/dt's just because it gives you the answer you want.
    That is not given in the OP.
    But it's dimensionally nonsense: you wrote h/T^2=b*l*2C/T^2*1/m
    In MLTQ notation, (mass, length, time, charge) LHS is LT-2.
    1T = 1Wb / 1m2; 1Wb = 1V · 1s; 1V = 1J / 1C; J has dimension ML2T-2. so V has dimension ML2T-2Q-1, Wb has dimension ML2T-1Q-1, Tesla has dimension MT-1Q-1. The RHS has dimension MT-1Q-1.L.Q.T-2.M-1 = LT-3
    It makes no sense to have an equation in which the two sides have different dimensions.
    Please post your school's solution. Either you've not understood it or your school needs some schooling!
  11. Nov 15, 2013 #10
    Ah yes ofcourse, youre right. It was F(t) not f'(t), thanks I made a silly error.

    I'm going to try it again using F(t), that should give me the right method. And yea I'll see if I find the solution.
  12. Nov 15, 2013 #11


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    One more thought - what you effectively calculated was the velocity in m/s. That makes sense: pushing that charge through the wire will have done work, which turned into KE. Are you sure this is not what the question asked for?
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