How Is the Magnetic Field Calculated to Levitate a Wire Carrying Current?

AI Thread Summary
To levitate a wire carrying a current, the magnetic field must produce an upward force equal to the weight of the wire. The required magnetic field direction is South, and its magnitude is calculated to be 0.102 T. The force on the wire is determined using the equation F = IL x B, where F equals the weight of the wire (mg). The discussion clarifies that 'L' in the equation refers to the length of the wire, not inductance. Understanding these principles allows for the calculation of the necessary magnetic field to achieve levitation.
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Homework Statement



A straight piece of wire of mass 0.0050 kg and length 0.600 m lies horizontally and carries an electrical current of I = 0.800 A in the direction E to W . What is the direction and magnitude of the magnetic field required to produce an upward force on the wire that exactly cancels its weight?

Homework Equations




The Attempt at a Solution



The direction is South and the magnitude of the magnetic field is 0.102 T. How did they calculate this? What principles and formulas are required here? :confused:
 
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Okay. Using this equation for force on a segment of current-carrying wire in a uniform magnetic fields F=ILxB, how would you relate this to the mass of the wire?
 
And how can I work out the inductance L? All the equations I know for L are for a solenoid...
 
You may have figured this already, but F = I (L x B) where F=ma & 'L' is length (not inductance)
 

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