A conducting wire ab of length l, resistance r and mass m starts sliding at t = 0 down a smooth, vertical, thick pair of connected rails as shown in . A uniform magnetic field B exists in the space in a diraction perpendicular to the plane of the rails. (a) Write the induced emf in the loop at an instant t when the speed of the wire is v. (b) what would be the magnitude and direction of the induced current in the wire? (c) Find the downward acceleration of the wire at this instant. (d) After sufficient time, the wire starts moving with a constant velocity. Find this velocity `v_m. (e) Find the velocity of the wire as a function of time. (f) Find the displacement of the wire as a functong of time. (g) Show that the rate of heat developed inte wire is equal to the rate at which the gravitational potential energy is decreased after steady state is reached.

Explanation

(a) when the speed the speed of wire is v
emf developed =Blv`
`(b) Induced current is the wire =(Blv)/R`
`(from b to a )`
`(c ) Downwards acceleration of the wire`
`=(mg-F)/m due to the current `
`=(ilB)/(m)=g=(B^2l^2v)/(Rm)`
`(d) let the start moving with constant velocity then acceleration =0`
`(B^2l^2v_m)/(Rm)=g`
` implies V_m=(GrM)/(B^2l^2)`
`(e) since (dv)/(dt)=a`
`implies (dv)/(dt)=((mg-B^2l^2v//R)/m) implies (dv)/((mg-B^2l^2v//R)/(m))=dt`
`int_(0)^(v)(m dv)/(mg-(B^2l^2v/R)/(m))=int_(0)^(v)dt`
`implies (m)/((B^2l^2v)/(R))[log (mg-((B^2l^2v)/(R))]_(0)^(v)=t`
` implies(-mR)/(B^2l^2)[log (mg-(B^2l^2v)/(R))-log(mg)]=t`
implies log[mg-((B^2l^2v)/(R))/(mg)]=(-t B^2l^2)/(mr)`
`=log[1-(B^2l^2v)/(Rmg)]=(-t B^2l^2)/(mr)`
` implies 1-(B^2l^2v)(Rmg)=e^((-t B^2l^2)/(mr)`
`implies (1-e^((-t B^2l^2)))=(B^(2)l^2v)/(Rmg)`
`v-(Rmg)/(B^2l^2)(1-e^((-t B^2l^2)))`
` implies v=v_m(1-e^(-(g t)/(v_m)))[:. v_m=(Rmg)/(B^2l^2)]`
` (f) since (ds)/(dt)=v`
`int ds= int v.dt`
`s=v_m(1-e^(-(g t)/(v_m)))`
`=v_m.(t+(v_m)/(g).e^(-(g t)/(vm)))`
`=(v_mt+(v_m)^2/(g)e^(-(g t)/(v_m)))-(V_m)^2/(g)`
`=v_m t=(V_m)^2/(g) (1-e^(-(g t)/(v_m)))`
(g) `(d)/(dt)(msg)=mg.(ds)/(dt)`
`=mg.(v_m)((1-e^(-(g t)/(v_m)))`
`(d_H)/(dt)=l^2R=R.((lBv)/(R))^2`
`=(l^2B^2v^2)/(R)`
` =(l^2 B^2)/(R).(V_m)^2((1-e^(-(g t)/(v_m)))^2`
`After steady state i.e..t rarr oo `
`(d)/(dt)(msg)=mg(v_m)`ltbr.`(d_H)/(dt)=(l^2B^2)/(R) (v_m)^2`
=(l^2B^2)/(r).(v_m).(mgR)/(l^2B^2)`
`Hence after steady state (d_H)/(dt)=(d)/(dt) mgs`.