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A 0.55 m copper rod with a mass of 5.00×10⁻² kg carries a current of 10 A in the positive x direction. Let upward be the positive y direction. What is the magnitude of the minimum magnetic field needed to levitate the rod?

User Janne
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Final answer:

The problem involves finding the minimum magnetic field needed to levitate a copper rod by balancing the gravitational and magnetic forces, using the formula for the magnetic force on a current-carrying wire.

Step-by-step explanation:

The question pertains to the concept of balancing gravitational and magnetic forces on a current-carrying wire to achieve levitation. Specifically, we have to find the minimum magnetic field required to levitate a copper rod with known mass, length, and current running through it. According to the formula for the magnetic force (Lorentz force) on a straight wire F = I × (L × B), where I is the current, L is the length vector of the wire, and B is the magnetic field vector, the minimum magnetic field will be that which counteracts the force of gravity, i.e., F = mg, with m being the mass of the rod and g being the acceleration due to gravity (approximately 9.81 m/s2). We can find this by rearranging the equation, solving for B, and then substituting the known values to find the magnitude of B.

User Dmitry Pavlov
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