A loop (figure) is formed by two parallel conductores connected by a solenoid with inductance `L` and a conducting rod of mass `m` which can freely ( without friction) slide over the conductors. The conductors are located in a horizontal plane in a uniform vertical magnetic field with induction `B`. The distance between the conductors is equal to `l`. At the moment `t=0` the rod is imparted an initial velocity `upsilon_(0)` directed to the right. Find the law of its motion `x(t)` if the electric resistance of the loop is negligible.

Explanation

We have in the circute at a certain instant of time `(T)`, from Faraday's law of electromagentice induction `:`
`L(di)/(dt)=Bl(dx)/(dt)` or `L di =B l dx`
As at ` t=0, x=0` so `Li=Blx` or `i=(Bl)/(L)x(1)`
For the rod from the second law of motion `F_(x)=mw_(x)`
` -ilB=mddot(x)`
Using Eqn. `(1)`, we get `:` `ddot(x) =-((l^(2)B^(2))/(mL))x=-omega_(0)^(2)x ....(2)`
where `omega_(0)lB//sqrt(mL)`
The solution of the above differential equation is of the form
`x=a sin (omegat+alpha)`
From the initial condition, at `t=0, x=0, ` so ` alpha=0`
Hence, `x=a sin omega_(0)t ....(3)`
Differentialting `w.r.t.` time, `dot (x) =a omega_(0)cos omega_(0)t `
But from the initial condition of the problem at `t=0, dot(x)=v_(0)`
Thus `v_(0)=a omega_(0)` or `a =v_(0)//omega_(0) ...(4)`
Putting the valur of `a` from Eqn `(4)` into Eqn. `(3)`, we obtained
`x=(v_(0))/(omega_(0))sin omega_(0)t(` where` omega_(0)=(lB)/(sqrt(mL)))`