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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
    sure, the magnetic field is going into the page, this is what the loop looks like, with the direction of the current and the magnetic field.

    [​IMG]
     
  5. Mar 14, 2006 #4

    nrqed

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    Ok. about Q1: the magnitude of the force is the magnitude of [itex] I \vec L \cross \vec B [/itex] or [itex] I L B sin \theta [/itex]. What is the value of the angle? (another way to see this is to look at the force on many small sections of the loop...you should see the answer quickly)

    About Q2: It looks like you used the length of the loop for the radius!

    Pat
     
  6. Mar 14, 2006 #5

    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!!!
     
  7. Mar 14, 2006 #6
    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.
     
  8. Mar 14, 2006 #7

    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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