two metal strips, each of length l are clamped parallel to each other on a horizontal floor with a separation b between them. A wire of mass m lies on them perpendicularly a s shown in the figure. A vertically upward magnetic field of strength B exists in the space. The metal strips are smooth but the coefficient of friction between the wire and the floor is u. A current i is established when the switch S is closed at the instant t=0. Discuss the motion of the wire after the switch is closed. How far away from the strips will the wire reach?
106F = Bib
So, a = Bib/m
So, it will undergo uniform acceleration till the wire reaches the end of the strips, after that it will undergo uniform decelration due to friction.
uniform acceleration till t=√2ml/iBb
speed of the wire till it reaches the end of the strips
v2 = 2(Bib/m)l
=> v = √2iBbl/m
Now, conserving energy,
mv2/2 = μmgx ... where x is the distance traveled by wire till it stops
=> iBbl = μmgx
=> x = iBblμmg
1thnx ashish....i didnt think that the wire woz allowed to side above the strips