Fe needed to lift up mass of size m

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btb4198
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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
um..
Am I doing this right? I feel like I am missing something...
 
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btb4198 said:
opposites magnetic fields repel each others.

no opposite poles attract :wink:
btb4198 said:
I = 3000 mA ( from a AA battery)

an AA battery is going to struggle to give 3A for any length of time ( other than some of your high current NiCd's)
btb4198 said:
um..
Am I doing this right? I feel like I am missing something...

don't really know, you showed a lot of formula without plugging in the numbers and showing your workingDave
 
in Fb = IL X B
what unit should L be in ?
meters?

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 ?