A 2.5 cm-long horizontal wire segment has mass 0.15 g and carries a current sufficiently large that it levitates when the segment is in a region of B field upward with magnitude 1.5 T.The current is passed to the segment that is in the magnetic field by way of massless electrical leads.What is the magnitude of the current that the wire carries?
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 upward and equal to the weight of the wire. The magnetic force is given by:$$F_m = I L B$$where:* $F_m$ 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 teslasThe 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 squaredSetting the magnetic force equal to the weight of the wire, we get:$$I L B = m g$$Solving for the current, we get:$$I = \frac{mg}{LB}$$Plugging in the given values, we get:$$I = \frac{(0.15 \times 10^{-3} \text{ kg})(9.81 \text{ m/s}^2)}{(0.025 \text{ m})(1.5 \text{ T})}$$$$I = 39 \text{ mA}$$

A 25 Cm-Long Horizontal Wire Segment Has Mass 0