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Fe needed to lift up mass of size m

  1. May 25, 2015 #1
    opposites magnetic fields repel each others. How big of an magnetic would I need to lift mass ( 5g) 3 centimetre in the air.
    F = MA
    A = 9.81 m/s^2
    M = mass of the object ( 5 g)
    F = 49.05

    F = I L x B
    49.05 = IL X B
    I = 3000 mA ( from a AA battery)
    µ = 4π E-7 T m/A
    Bsol = µ N/L I
    F = IL X µ N/ L * I
    N= number of turns in coil

    N = 10
    L = 10 m
    Am I doing this right? I feel like I am missing something...
  2. jcsd
  3. May 25, 2015 #2
    Mass of object must be kilogram.
  4. May 25, 2015 #3


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    no opposite poles attract :wink:

    an AA battery is going to struggle to give 3A for any length of time ( other than some of your high current NiCd's)

    don't really know, you showed a lot of formula without plugging in the numbers and showing your working

  5. May 27, 2015 #4
    in Fb = IL X B
    what unit should L be in ?

    if F = 49.05 N that mean that you need Fb to be greater than 49.05 to get the object off the ground right?
    but how do you get how high up it will go ?
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