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!

Dragging loop out of magnetic field (Finding Force)

  1. May 17, 2009 #1
    1. The problem statement, all variables and given/known data
    A single rectangular loop of wire with width w = 30.7 cm and length L = 79.7 cm is situated so that part is inside a region of uniform magnetic field of 0.407 T and part is outside the field. The total resistance of the loop is 0.244 Ω. Calculate the force F required to pull the loop from the field (to the right) at a constant velocity of 3.56 m/s. Neglect gravity.
    HELP: What current is induced when the loop is pulled from the field? What is the magnetic force on the current-carrying wire of the loop?

    https://wug-s.physics.uiuc.edu/cgi/courses/shell/common/showme.pl?cc/Knox/phys130a/spring/homework/15/02/HW16_5.jpg [Broken]

    2. Relevant equations
    emf= -N([tex]\Delta[/tex]BA/[tex]\Delta[/tex]t)
    I= emf/R
    F= NILB


    3. The attempt at a solution
    I found the emf to be ..448 (-1(0-(.407x.3985x.309))/.1119=.448)
    From there I found the current to be 1.836 A and using that I found the Force to be .298
    (1.836x.3985x.407=.298)
    This is incorrect and I'm not quite sure what I'm doing wrong. The use of the velocity throws me off and I have a feeling my mistake is coming from there.
     
    Last edited by a moderator: May 4, 2017
  2. jcsd
  3. May 17, 2009 #2

    nrqed

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    I am not sure how you got the numbers 0.3985 and .309..

    Here, for delta A/delta t you may use

    [tex] \frac{\Delta A}{\Delta t} = \frac{w \Delta L}{\Delta t} = v w [/tex]

    Then the change of flux is just B v w
     
    Last edited by a moderator: May 4, 2017
  4. May 17, 2009 #3
    ... .309 was supposed to be .307, so that's my mistake. And then.3985 was half of .797 for the length.
    The change of flux is the top part of the equation for my emf, right? In which case, the time is still just 1, and that would be equal to my emf. If this is the case, I'm still getting the answer wrong.
    But I think I'm just not understanding where the change of flux comes into play.
     
  5. May 17, 2009 #4

    nrqed

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    Ok, I see what you did.
    The drawing seems to show the situation when half the loop is in the magnetic field, but we don't need to assume this to do the problem.

    You can do it by looking at the loop at two different times. Then what you must do is to calculate the change of area of the part of the loop submerged in the magnetic field.

    Let's consider the situation at some intitial time t_i and then one second later, ok?
    In that one second, the loop has moved a distance

    [tex]v t = 3.56 m/s \times 1 s = 3.56 m [/tex]

    so the part of the length submerged in the magnetic field has decreased by 3.56 m. Since w =0.307 m, the area submerged in the magnetic field has decreased by 3.56 m times w = 1.093 m^2. This is the change of area. Then multiply this by the B field to find the change of flux.
     
  6. May 17, 2009 #5
    Oh, wow, that worked perfectly. Thanks a bunch!
     
  7. May 17, 2009 #6

    nrqed

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    You are very welcome:smile:
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook