Levitating a rod - magnetic field/forces

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To levitate a 2.0 g wire, the magnetic field strength must be calculated using the formula B = F/(IL). The force (F) acting on the wire is determined by its weight, which is 2 g converted to kg (0.002 kg) multiplied by gravitational acceleration (9.8 m/s²), resulting in F = 0.0196 N. The current (I) is given as 1.5 A, and the length (L) is 0.10 m, leading to IL = 0.15. The correct calculation for B results in approximately 130.67 T, but the initial attempt was incorrect due to not converting grams to kilograms. Accurate unit conversion is essential for solving this problem correctly.
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Homework Statement



What magnetic field strength will levitate the 2.0 g wire in the figure?

knight_Figure_32_36.jpg



Homework Equations



F = ILB


The Attempt at a Solution



Solving for B, the formula becomes:

B = F/(IL)

F = 2*9.8 = 19.6
IL = (1.5)*(0.10) = 0.15
B= 19.6/0.015 = 130.666666666...

which comes out as incorrect

what am i doing wrong?

thanks!
 
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...anybody know?

lol i have to submit this in a few hours! :(
 
Last edited:
convert the 2g to kg
 
i was thinking of doing that...but i wasnt sure..but yea that is correct

THANKS!
 

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