A coil of inductance `L` connects the upper ends of two vertical copper bars separated by a distance `l` . A horizontal conducting connnector of mass `m` starts falling with zero initial velocit along the bars without losing contact with them. The whole system is located in a uniform magnetic field with induction `B` perpendicular to the plane of the bars. Find the law of motion `x(t)` of the connector.

Explanation

As the connector moves, an emf is set up in the cirucuit and a current flows, since the emf is `xi=-Bl dot (x), ` we must have `: -Bl dot (x)+L(dI)/(dt)=0`
so, `I=Blx//L`
provided `x` is measured from the initial position.
We then have
`m ddot (x)=-(Blx)/(L). B. l + mg`
for by Lenz's law the induced current will oppose downward sliding . Final
`ddot(x)+((Bl)^(2))/(mL)x=g`
on putting `omega_(0)=(Bl)/(sqrt(mL))`
` ddot(x) + omega_(0)^(2)x =g`
A solution of this equation is `x=(g)/(omega_(0)^(2))+A cos (omega_(0)t+alpha)`
But` x=0` and `dot(x) =0` at `t=0.` This gives
`x=(g)/(omega_(0)^(2))(1-cos omega_(0)t)`.