A 20 cm long conducting wire of 5 g carrying 5 A current is placed in a uniform magnetic field. The strength and direction of the magnetic field needed to levitate the wire are

Question

darkknight · Answer

Answer choices? Are there any?

Vocaloid · Answer

force exerted on the wire due to the magnetic field: F = BIL where B is the strength of the magnetic field, I is current, L is length (be sure to convert to SI units)

in order to levitate, the force from the magnetic field must be equal to the force of gravity. F = mg, again being sure to convert to SI units first.

BIL = mg, solve for B to get the magnitude of the field, and then use RHR to determine the direction