Two long parallel wires of zero resistance are connected to each other by a battery of `1.0V` . The separation between the wires is `0.5m` . A metallic bar, which is perpendicular toi the wire and of resistance `10Omega` moves on these wire when a magnetic field of `0.02` tesla is acting p endicular to the plane containing the wire and the wires. Find the velocity of the bar as a function of time if the mass of the bar is `0.002kg` . Find also the steady-state velocity of the bar.

Explanation

Due to current `i` supplied by batery, force experienced by bar,
`F=Bil=(Bel)/(R)(` :. `i=(E)/(R),R` : resistance of rod) Due to force `F` , rod accelerates towards right.
Let at any time `t` , velocity of rod is `v` .
Induced emf `e=Bvl`
`F'=Bi'l=(B(E-Bvl)l)/(R)=ma`
Acceleration of rod `a=(Bl(E-Bvl))/(mR)`
`(dv)/(dt)=(Bl)/(mR)(E-Bvl)`
`int_(0)^(v)(dv)/(E-Bvl)=(Bl)/(mR)int_(0)^(t)dt`
`(|1n(E-Bvl|_(0)^(v)))/(-Bl)=(Bl)/(mR)t`
`1n(E-Bvl)-1n(E)=-(B^(2)l^(2))/(mR)`
`(E-Bvl)/(E)=e^(-B^(2)l^(2)t)/(._emR)`
`v=(E)/(Bl)(._(1-e)(-B^(2)l^(2)t)/(Mr))`
`=(1)/(0.02xx0.5)[1-e((0.02)^(2)(0.5)^(2)t)/((0.002)(10))]`
`v=100(1-e^(-(t)/(200)))`
At `t=oo` , `v=v_(T)=100m//sec` OR At steady-state velocity, `a=0`
`(Bl(E-Bvl))/(mR)=0`
`v=v^(T)=(E)/(Bl)=(1)/(0.02xx5)=100m//sec`