Question

a long, horizontal wire ab rests on the surface of a table and carries a current i. horizontal wire cd is vertically above wire ab and is free to slide up and down on the two vertical metal guides c and d (fig. e28.27). wire cd is connected through the sliding contacts to another wire that also carries a current i, opposite in direction to the current in wire ab. the mass per unit length of the wire cd is l. to what equilibrium height h will the wire cd rise, assuming that the magnetic force on it is due entirely to the current in the wire ab?

Answer 1

The wire cd will rise to an equilibrium height h due to the magnetic force exerted on it by the current in wire ab.

Step-by-step explanation:

The wire cd will rise to an equilibrium height h due to the magnetic force exerted on it by the current in wire ab. The magnetic force on wire cd is given by the equation F = iLB, where i is the current in wire ab, L is the length of wire cd, and B is the magnetic field strength. The force exerted by the weight of wire cd is given by the equation F = mg, where m is the mass per unit length of wire cd and g is the acceleration due to gravity.

At equilibrium, the magnetic force is equal to the weight force, so we can set the two equations equal to each other:

iLB = mg

From this equation, we can solve for the equilibrium height h:

h = L - (mg)/(iB)