Magnetic fields and current flows

AI Thread Summary
To levitate a 2.50-meter long copper rod with a mass of 0.25 kilograms in a 0.15 Tesla magnetic field, a minimum current of approximately 6.533 A is required, flowing in the east direction. The calculations are based on the equations F = mg and F = ILB, leading to the formula I = mg/(LB). The discussion highlights the importance of understanding the underlying concepts and formulas in physics. Participants encourage perseverance and confidence in mastering these topics. Mastery of these principles is essential for solving similar problems in electromagnetism.
liz_p88
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Homework Statement



A 2.50 meter long copper rod has a mass of 0.25 kilograms. A magnetic field of 0.15 Teslas is oriented horizontally from south to north. a.) What is the minimum current in the rod that would allow it to "levitate" in this magnetic field? b.) What is the direction of the current in the wire?

Homework Equations



F = mg
F = ILB
ILB = mg

The Attempt at a Solution



I = mg/(LB)

(.25kg)(9.8m/s^2)/(2.5m)(.15T) = 6.533 A, East

I'm not sure if I did this correctly
 
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yes you are correct.
 
Thank you! I doubt myself so much. I have to work so hard to understand these concepts and formulas
 
yes, work hard and you will gain confidence.
 

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