The rectangualr wire- frame, shown in has a width d, mass m, resistance R and a large length. A uniform magnetic field B exists to the left of the frame. A constant force F starts pushing the frame into the magnetic field at t =0. (a) Find the acceleration of the frame when its speed has increased to v. (b) Show that after some time the frame will move with a constant velocity till the whole frame enters into the magnetic field. find this velocity `v_0`. (c) show that the velocity at tiem t is given by ` v= v_0(1 - e^(-Ft/mv_0)`.

Explanation

e.m.f developed = Bdv
(when it attains a speed v)
`Current =(Blv)/(R )`
` this force opposes the given force`
`Net F=F-(Bd^2 v^2)/(R ) (RF-Bd^2v^2)/(R )`
` net acceleration =(RF-Bd^2v^2)/(mR)`
`(b) Velocity becomes constant when acceleration is 0.`
`(F/m)-(Bd^2v^2)/(mR) [V=V_0 at that instant]`
`=(F/m)-(Bd^2v^2)/(mR)`
` implies V_0= (FR)/(B^2d^2)`
`(c ) Velocity at time `
` :. a=(dv)/(dt)`
` implies (Rf-l^2B^2v^2)/(mR)=(dv)/(dt)`
` implies int_(0)^(v) (dv)/(RF-l^2B^2v) = int_(0)^(t) (dt)/(mR)`
` implies [In(RF-l^2B^2v)]_(0)^(v) =[(t)/(Rm)]_(0)^(t)-In (RF)=(l^2B^2t)/(Rm)`
` implies (l^2B^2v)/(RF)=1-e^(l^2B^2v)/(Rm)`
` v= (FR)/(l^2B^2)(1-e^(B^2l^2v_0t)/(Rv_)m))`
`=v_0(1-e^(-Ft/v_0m)`.