What is the Direction of the Magnetic Field Needed to Levitate a Copper Rod?

AI Thread Summary
To levitate a copper rod carrying a current of 10 A in the positive x direction, a magnetic field of 0.26 T is required. The direction of the magnetic field needed for levitation is determined using the right-hand rule. Given that the current flows to the right (positive x) and the required force for levitation is upward (positive y), the magnetic field must be directed in the positive z direction. This aligns with the principles of electromagnetism and the interaction between current and magnetic fields. Understanding these concepts is crucial for solving similar physics problems.
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Homework Statement


A 0.49 m copper rod with a mass of 0.13 kg carries a current of 10 A in the positive x direction. Let upward be the positive y direction.
1) What is the magnitude of the minimum magnetic field needed to levitate the rod?
2) What is the direction of the minimum magnetic field needed to levitate the rod?
pos x direction, neg x, pos y, neg y, pos z, neg z

Homework Equations



Right Hand Rule?

The Attempt at a Solution


I figured out part 1 but not sure on part two. any guidance would be appreciated. I'm thinking it's positive y direction. Part 1 answer was B = .26 T fwiw
 
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Yup, simply use the right hand rule. You know the current is going to the right (positive x) and to levitate the rod you would need a force going up (positive y). So by using the right hand rule you should be able to find the direction of the magnetic field.
 
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