Physics 102?

Question

A unusual messege delivery system is pictured . A 15-Cm length of conductor that is free to move is held in place between two thin conductors. When a 5 -A current is directed the wire segment moves upward at a constant velocity. IF the mass of the wire is 0.15 kg find the magnitude and direction of the minumum magnetic field required to move the wire. ( the wire slides without friction on the two vertical conductors.)

Answer 1

Start with F = I*L*B  -->  B = F/(I*L) = (m*g)/(I*L)

m = 0.15kg

g = 9.8m/s²

I = 5A

L = 0.15m

Substitute and solve to find magnitude of B.

The force F needed to move the wire must be directed opposite the force of gravity, thus F must be directed up.  To find direction of B, use the right-hand-rule:  thumb=F, index=I, middle=B.  B will be into the page if I is directed L-->R, or out of the page if I is directed R-->L.