Fe needed to lift up mass of size m

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Discussion Overview

The discussion revolves around the calculations needed to determine the strength of a magnetic field required to lift a mass of 5 grams a distance of 3 centimeters using electromagnetic principles. Participants explore the relevant formulas and units involved in this context.

Discussion Character

  • Technical explanation
  • Mathematical reasoning
  • Debate/contested

Main Points Raised

  • One participant presents a calculation using the formula F = MA to determine the force needed to lift a 5g mass, suggesting a force of 49.05 N is required.
  • Another participant points out that the mass should be converted to kilograms for the calculations to be correct.
  • Concerns are raised about the current output of an AA battery, with a participant noting that it may not sustain 3A for long periods.
  • Questions are posed regarding the appropriate units for the length (L) in the formula Fb = IL x B, and whether it should be in meters.
  • There is a query about the relationship between the force needed to lift the object and how to calculate the height it will rise.

Areas of Agreement / Disagreement

Participants express uncertainty about the calculations and units, indicating that there is no consensus on the correct approach or the assumptions made in the calculations.

Contextual Notes

Limitations include the need for unit conversions, the dependence on the specifications of the battery, and the lack of numerical examples in the calculations presented.

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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Mass of object must be kilogram.
 
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 ?
 

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