A loop is formed by two parallel conductors 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 magnet field `B` . the distance between the conductors is `l` . At the moment `t=0` , the rod is imparted an initial velocity `v_(0)` directed to the right. Find the law of its motion `x(t)` if the electric resistance of the loop is negligible.

Explanation

Let at any `t` , velocity of rod is `v` , motional emf `e=Bvl`
`L(di)/(dt)=Bvl=Bl(dx)/(dt)`
`Ldi=Bldx`
Integrating
`Li=Blx implies i=(Bl)/(L)x`
Magnetic force on rod
`F_(m)=Bil=(B^(2)l^(2))/(L)x` , Force opposes motion
`ma=(B^(2)L^(2))/(L)x`
`a=(B^(2)L^(2))/(mL)x`
`(d^(2)x)/(dt^(2))=-(B^(2)l^(2))/(mL)x`
`=-omega^(2)x` (where `omega=(Bl)/(sqrt(mL)))`
This is the equation of `S.H.M` . The rod osicillates simple harmonically with angular frequency `omega=Bl//sqrt(mL)` .
`(d^(2)x)/(dt^(2))=-omega^(2)x`
The solution of this equation of the type
`x=Asin(omegat+phi)`
where `A` and `phi` be determined by initial conditions
`t=0` , `x=0` , `phi=0`
`x=Asinomegat`
`(dx)/(dt)=Aomegacosomegat`
`v=Aomegacosomegat`
At `t=0` , `v=v_(0)`
`v_(0)=Aomega implies A=(v_(0))/(omega)`
`x=Asinomegat`
`x=(v_(0))/(omega)sinomegat` , where `omega=(Bl)/(sqrt(mL))` OR When velocity of rod `v` and current in the rod is `i` , by energy conservation (no resistance, no loss energy)
`(1)/(2)mv^(2)+(1)/(2)Li^(2)=(1)/(2)mv^(0^(2))`
`i=sqrt((m)/(L))sqrt(v_(0)^(2)-^(2))` ...(i)
Force on rod, force opposes the motion
`F_(m)=Bil`
`ma=-Bil`
`a=-(Bl)/(m)i=-(Bl)/(m)sqrt((m)/(L))sqrt(v_(0)^(2)-v_(0))`
`(dv)/(dt)=-(Bl)/(sqrt(mL))sqrt(v_(0)^(2)-v_(0))`
`(dv)/(dt)=-omegasqrt(v_(0)^(2))(omega=(Bl)/(sqrt(mL)))`
`int_(vo)^(v)(dv)/(sqrt(v_(0)^(2)-v^(2)))=-omegaint_(0)^(t)dt`
`|sin^(-1)((v)/(v_(0)))|_(v_(0))^(v)=-omegat`
`sin^(-1)((v)/(v_(0)))-sin^(-1)((v_(0))/(v_(0)))=-omegat`
`sin^(-1)((v)/(v_(0)))-(pi)/(2)=-omegat`
`sin^(-1)((v)/(v_(0)))=(pi)/(2)-omegat`
`(v)/(v_(0))=sin((pi)/(2)-omegat)=cosomegat`
`v=v_(0)cosomegat`
`(dx)/(dt)=v_(0)cosomegat`
`int_(0)^(x)dx=v_(0)int_(0)^(t)cosomegatdt`
`x=v_(0)(|sinomegat|_(0)^(t))/(omega)=v_(0)/(omega)sinomegat`
`x=v_(0)/(omega)sinomegat`
where `omega=(Bl)/(sqrt(mL))`
The rod executes simple harmonic motion of amplitude `(v_(0))/(omega)` and angular frequency `(Bl)/(sqrt(mL))`