1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: The speed of a metal wire on two rails with a magnetic field

  1. Sep 9, 2018 #1
    1. The problem statement, all variables and given/known data
    I tried to understand the problem b) and c).

    upload_2018-9-9_17-11-32.png

    2. Relevant equations
    Faraday's law: ∇xE = - ∂B/∂t
    emf ε = Bdv
    Force : F =ma, Lorenz's force F=q(vxB) ==> ma = IdB
    Power : power of battery = εI, mechanical power of the wire = Fv

    3. The attempt at a solution
    I think I solved a), where the magnetic force on the wire is F = IdB = ma and the speed v = at = IdBt/m.
    I'm now tried to understand b) and c). First, I assume the initial speed is 0 at t=0 when the generator is replaced by a battery. The battery's power is εI, and it must be equal to the power consumed by the wire which is Fv = IdBv. Then the speed of the wire is v = ε/dB.
    Also using Faraday's law I get to the same conclusion : ∇xE = - ∂B/∂t >>> ε = -∂(flux)/∂t = Bdv
    It doesn't feel natural. The battery is connected at t=0, and in no time there the wire goes at speed v=ε/dB? So I think it is nonsense. But I don't know where I choose a wrong way. Help me plz.
    I guess the answer to c) is that the current has to be 0, because when the terminal speed has been reached, there is no acceleration, then no force F=IdB=0. Am I right at this?
     
  2. jcsd
  3. Sep 9, 2018 #2

    kuruman

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    In part (a), you cannot use ##v=at## because the acceleration is not constant. As the bar accelerates, the current ##I## is also not constant, but depends on the velocity which depends on time. You need to solve a differential equation for part (a). Start with Newton's 2nd Law in the form $$m\frac{dv}{dt}=IB~d$$and then use Faraday's Law to find ##I## in terms of ##v##. Put that back in Newton's 2nd Law and solve.
    How do you figure? The variable "v" cancels out in the equation.

    On edit: Ooops! I meant part (b) in what I wrote above, not part (a). Part (a) is correct because the rails are connected to a constant current generator.
     
    Last edited: Sep 10, 2018
  4. Sep 10, 2018 #3
    Perhaps this thread is similar to (b) and (c) questions in your problem:
    https://www.physicsforums.com/threa...-inside-a-magnetic-field.952728/#post-6036839
    In your problem (questions (b) anc (c)) there are a force due the constant potential ##\varepsilon##. In the above link the force is due to a constant gravitational potential near the Earth surface. Is a little bit different but I think above thread can help you.

    kuruman, I've have been thinking about this problem from yesterday because it is hard. I don't know if in the (a) question there are a variable electric potential that compensates ##\varepsilon_inducc## and then Intensity really is a constant. I don't know.
    Edit.
    Also it is important the direction of circulation of Intensity, because the bar must be moved to the left in the picture.
     
  5. Sep 10, 2018 #4

    kuruman

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    I retracted my statement on edit. Part (a) is correct.
    Why? The direction of the field is shown in the figure, but without knowing which way the battery is connected, one cannot figure out the direction of the velocity.
     
  6. Sep 10, 2018 #5
    I says about the direction o velocity because the diagram, if the bar moves to the right it will crashes whit the generator. Ok. if the supports wires are so long this is not important.

    knowone needs to calculate ##\varepsilon_{inducc}## that is related with ##v## (when he applies Faraday's Induction law). After that I think that he can try:
    $$\varepsilon -\varepsilon_{inducc}=I·R$$
    Then knowone can get Intensity, and put it into the movement equation that you has wrote above.
    Then he will need to solve the differential equation.
    Please if there are some mistake say me it. In my work problems do not are so hard.
     
  7. Sep 10, 2018 #6

    kuruman

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    Your reasoning is correct. Let's wait and hear what @knowone has to say.
     
  8. Sep 16, 2018 at 1:32 AM #7
    The power of the battery is εI and it equals Fv=IdBv. Thus εI=idBv and v= ε/dB
     
  9. Sep 16, 2018 at 2:07 AM #8
    I solved these equations:
    εbatteryinduced =IR, εinduced = vdB, and m(dv/dt)=IBd
    Then I= (εbattery-vdB)/R, thus m(dv/dt)=Bd((εbattery-vdB)/R).

    Using differential equation, v(t)=A(1-e-λt), where A = -εbattery/d2B2 and λ = d2B2/mR

    Here we added a arbitrary constant R. I'm reluctant to use it because there is no specification about any resistance of the circuit.

    If we are allowed to use the solution v(t), then what will be the time when the system arrives at equilibrium?
     
    Last edited: Sep 16, 2018 at 2:16 AM
  10. Sep 16, 2018 at 6:59 AM #9

    kuruman

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    You do need R to express v(t). I believe it should have been given by the problem. Your expression for v(t) = A(1-e-λt) is correct. Clearly A is the terminal speed which should be positive. You can see that from the differential equation. At terminal speed, dv/dt = 0. What do you get when you solve for v on the right side to get the terminal speed?
    The time will be mathematically infinite but the problem is not asking for it.
     
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted