A 2.5 cm-long horizontal wire segment has a mass of 0.15 g and carries a current sufficiently large that it levitates when the segment is in a region of uniform B field of magnitude 1.5 T. Assume that the B field is horizontal and perpendicular to the wire segment. The current is passed to the segment by way of massless electrical leads. What is the magnitude of the current in the wire?
A) 39 mA
B) 25 A
C) 4.0 mA
D) Not enough information is given.

Question

Quizplus · Accepted Answer

The magnetic force on the wire is given by:$$F_B = I L B$$where:* $$F_B$$ is the magnetic force (in newtons)* $$I$$ is the current (in amperes)* $$L$$ is the length of the wire (in meters)* $$B$$ is the magnetic field strength (in teslas)The weight of the wire is given by:$$F_g = m g$$where:* $$F_g$$ is the weight of the wire (in newtons)* $$m$$ is the mass of the wire (in kilograms)* $$g$$ is the acceleration due to gravity (in meters per second squared)In order for the wire to levitate, the magnetic force must be equal to the weight of the wire:$$F_B = F_g$$Substituting the given values into the equations above, we get:$$I L B = m g$$$$I = \frac{mg}{LB}$$$$I = \frac{(0.15 	imes 10^{-3} 	ext{ kg})(9.81 	ext{ m/s}^2)}{(0.025 	ext{ m})(1.5 	ext{ T})}$$$$I = 39 	ext{ mA}$$Therefore, the magnitude of the current in the wire is 39 mA.

A 25 Cm-Long Horizontal Wire Segment Has a Mass of 0