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Electrodynamics problem(Lenz' s Law)

  1. Jan 22, 2004 #1
    A metal bar of mass m slides frictionlessly on two conducting rails a distance l apart as shown in the attached Figure 1. A resistor R is connected across the rails and a uniform magnetic field B, directed into the page fills the entire region.

    a) If the bar is moving to the right at speed v, what is the current flowing through the resistor?
    b) In what direction is the current flowing?
    c) Calculate the magnetic force acting on the moving bar.
    d) What is the direction of the magnetic force acting on the moving bar?
    e) If the bar is given an initial speed v0 at time t = 0 directed towards the right, and is thereafter left to slide on its own, derive an expression for the speed of the bar at time t > 0.
    f) The initial kinetic energy of the bar was 1/2mvo^2. If the bar is alllowed to slide until it comes to a stop, show that the energy dissipated in the resistor is 1/2mv0^2.

    I don' t know if the picture is goina come out right...there is a large rectangular loop with a resistor along the left side of it in the center. The right side is removed and somewhere along , there is a bar parallel to the left side that is being moved.
    | ||
    | ||
    | ||
    R || <----- bar of mass m
    | ||
    | ||
    | ||
    the flux through the resistor is increasing as the bar is moved to the right. This means that there should be an emf that causes a magnetic field that has to oppose the change in flux. This induced B field should be pointing out of the page. This should mean that the current is flowing counter clockwise. I am not sure how to calculate this current.

    For the magnetic force I know that it is supposed to be some kinda integral but not sure what.... For its direction, well would this direction be dependent on the result of the integral.. I mean after choosing a path and getting a non zero result, if the result is negative then the path has to be traversed in the opposite direction and then the direction is easily seen by just flipping the direction.

    Now, for 3 and f, I am really lost. I have not seen any speed or energy derivation related to this kind of thing in the text so I am really confused as to what I would be comparing it to...I think it might be something like E before = E after but do not know the elements.

    Any assistance would be great..thanks!
    Last edited: Jan 22, 2004
  2. jcsd
  3. Jan 22, 2004 #2
    You need to use Faraday's Law of Induction to find the current:

    Induced EMF = the change in magnetic flux with respect to time.

    The flux through the area of the circuit = Blx (easy, because the magnetic field is uniform.)

    l = length of rod.
    x = the distance between the bar and the resistor.

    In this case x is changing with time. (dx/dt)

    Since the magnetic field is uniform all we have to worry about is dx/dt which is v, the velocity.

    So now the induced EMF = Blv.

    You should be able to solve for the current now and you've already figured out the direction.
  4. Jan 22, 2004 #3
    Hey man thanks a lot...it really helped me out..I understood what was going on very clearly .

    Here' s what I get as the final solutions:
    a) I took EMF =-d(Phi)/dt and then got EMF = -Blv.
    b) Current is counter clockwise and I = -Blv/R
    c) F = iL X B and got that F = IlBsin90 = -B(l^2)v/R
    d) the force points along the vertical segement of the loop and is directed downwards.

    Do these seem correct to you?
  5. Jan 22, 2004 #4
    Also a few more questions:

    1) Parts e and f..are very puzzling to me ..can someone help me out the same way you did with parts a through to d ?

    2) A square loop with sides of length a is mounted on a shaft and rotated vertically at an angular speed omega in the prescence of a uniform magnetic field B. Derive an expression for the EMF induced in the loop. I have solved this question for the case where there were N loops of sides a and the rotational was along the horizontal axis i.e. the loope was rotated horizontally. Would the solution be the same except that N would have to be made to equal 1 in this case?

    Thanks for the help and and looking forward to anyone' s helpful replies to these ones.
    Last edited: Jan 22, 2004
  6. Jan 22, 2004 #5
    Ok, for the question relating to the rotating loop. I looked up the question that I had solved and the one I have to solve and both mention that the apparatus is that of an alternating current generator..I guess that means that the rotation about a horizontal or vertical shaft will not make a difference right? It seems that way to me .. am I mistaken?
  7. Jan 22, 2004 #6
    Come on..I am sure that this is not too difficult for someone to help me with.
  8. Jan 22, 2004 #7
    c) F = -(B^2)(l^2)v/R
    d) the right hand rule for iL X B yields a Force in the plane of the page to the left, since the induced current runs counterclockwise.
  9. Jan 22, 2004 #8
    Start with Newtons Second Law:

    F = ma = -(B^2)(l^2)v/R ==> m(dv/dt) = -(B^2)(l^2)v/R

    manipulate the equation and you'll end up with the two integrals you're looking for with the initial conditions v = v0 and t = 0.

    Can you take it from here?
  10. Jan 22, 2004 #9
    Oh yeah thanks,,,but shouldn' t the current be tothe right..iL X B for B into the page is to the left right?
  11. Jan 22, 2004 #10
    Would these be the integrands..... m dv / v = - B^2 * I ^ 2 * dt / R and the integral runs from vo to v and t= 0 to t = t right?
  12. Jan 22, 2004 #11
    i is running counterclockwise through the circuit and L is the length of the bar, so you're considering the current running through the bar which in this case runs from the bottom of the bar to the top of the bar.

    If you apply the right hand rule and cross the current running up the page with B going into the page, your thumb is pointing to the left.
  13. Jan 22, 2004 #12
  14. Jan 22, 2004 #13
    I have an idea for f)...

    Energy at t = 0 = Energy at t = t(when the bar stops)

    1) Energy at t = 0 is just (1/2)*m*v0^2 + zero energy in the resistor
    2) Energy at t = t is just (1/2)*m*0^2 + energy that appers in the resistor.

    Thus by setting 1 = 2 I would get that the energy that appears in theresistor is the given 1/2)*m*v0^2....

    The only problem I have with this is that it just seems too easy...Is the explanation I have given above for f) correct?
  15. Jan 23, 2004 #14
    Hey thanks, I finally figured out the right hand rule

    1) What do you guys think of my explanation for part f) ? IS it correct?

    2) Also what do you guys think of the alternating generator question...does the shaft about which rotation is occuring have to be horizontal or can it be vertical?
  16. Jan 24, 2004 #15
    cOME ON GUYS...I depsarately need help with aprt f)..any suggestions would be breat.
  17. Jan 24, 2004 #16
    Yes energy is conserved:

    KE0 + U0 = KEF + UF
    UF = KE0
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