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Homework Help: Magnetic Force,EMF, Flux question

  1. Mar 14, 2006 #1
    I have three questions that I think I have the answers to, but would like someone to tell me if I did them correctly. Thanks!
    Q1. Given B= 0.8T, a coil with 1 loop with length of 20cm, a current of 5.0A, what are the size and direction of the magnetic force on the loop?
    A1. Would I just use Fm= IxBxL = (5A)(.8T)(.2m) = .8 N ???
    As for the direction I know it involves the right hand rule, and the current moves to the right, so would the magnetic force be pointing up ?

    Q2. What is the magnetic flux through the loop?
    A2. Would I use: Flux= ABcos0 = pi(.2)^2 (.8T) (cos0) = .101 Wb ??

    Q3. If the magnetic field has a cut off at the top of the loop, and the loop is pulled completely out of the field upwards at v = .2 m/s:
    what EMF would be generated and if the resistance was 4 ohms, what would the current be?
    A3: EMF= Blv = (.8T)(.2m)(.2m/s) = .032V
    O = EMF/R = .032V / 4 ohms = .008A


    do these look right? thanks.
     
  2. jcsd
  3. Mar 14, 2006 #2

    nrqed

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    You have to tell us in what direction the B field is pointing!!! (with respect to the normal of the loop, let's say)
     
  4. Mar 14, 2006 #3

    nrqed

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    Oh wait, it's a *square* loop (when I saw your solution I thought you had in mind a circular loop!). Then it's ok to square the length of one side, but there shoul dbe no factor of [itex] \pi [/itex] there!!!
     
  5. Mar 14, 2006 #4
    whoops, two things: i forgot to mention that in question 1 I nedd to find the magnetic force and direction on just the Lower section of the loop, and the second thing is i guess i wasnt thinking about the pi, thanks for pointing that mistake out.
    If i am looking at just the lower section of the loop would it be: F= IBLSin(theta) where theta is 90??
    thanks.
     
  6. Mar 14, 2006 #5

    nrqed

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    Ah.. ok. Yes, then theta is 90. I assume you know how to get the direction of the force...

    And your answer for the induced emf seems correct.

    Pat
     
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