Consider the situation shown in the figure. The wire `PQ` has mass `m` , resistance `r` and can slide 0n the smooth, horizontal parallel rails separated by a distance `l` . The resistance of the rais is negiligible. A uniform magnetic field `B` exists in the reactangular region and a resistance `R` connects the rail outside the field region. At `t=0` , the wire `PQ` is pushed toward right with a speed `v_(0)` . Find
(a) the current in the loop at an instant when the speed of the wire `PQ` is `v`
(b) the acceleration of the wire as this instant
(c) the velocity `v` as a function of `x`
(d) the maximum distance the wire will move
(e) the velocity as a function of time

Explanation

When speed of wire `PQ` is `v` , it can be replaced by a battery of emf `e=Bvl`
(a)`I=(e)/(R+r)=(Bvl)/(R+r)`
Force on wire `PQ` ,
`F=Bil=(B^(2)vl^(2))/(R+r)` , towards left
(b) Retardation of wire `a=(B^(2)l^(2))/(m(R+r))`
(c) `a=(B^(2)l^(2))/(m(R+r)v`
`v(dv)/(dx)=-(B^(2)l^(2))/(m(R+r)v`
`int_(v0)^(v)dv=-(B^(2)l^(2))/(m(R+r)int_(0)^(x)dx`
`|v|_(v0)^(v)=-(B^(2)l^(2))/(m(R+r)|x|_(0)^(x)`
`v-v_(0)=-(B^(2)l^(2))/(m(R+r)x`
`v-v_(0)=-(B^(2)l^(2))/(m(R+r)`
(d) The wire will stop, when `v=0`
`x_(0)=(mv_(0)(R+r))/(B^(2)l^(2))`
(e) `a=(B^(2)l^(2))/(m(R+r))v=lambdav`
where `lambda=(B^(2)l^(2))/(m(R+r))`
`(dv)/(dt)=-lambdav`
`int_(v0)^(v)(dv)/(v)=-lambdaint_(0)^(t)dt`
`|1nv|_(vo)^(v)=-lambdat`
`1nv-1nv_(0)=-lambdat`
`1n(v//v_(0))=-lambdat`
`(v)/(v_(0))=e^(-lambdat)`
`v=v_(0)e^(-lambdat)`