A metal bar AB can slide on two parallel thick metallic rails separated by a distance l. A resistance R and an inductance L are connected to the rails as shown in the figure. A long straight wire carrying a constant current `I_(0)` is placed in the plane of the rails and perpendicular to them as shown. The bar AB is held at rest at a distance `x_(0)` from the long wire. At t=0, it is made to slide on the rails away from wire. Answer the following questions.
(a) Find a relation among `i, (di)/(dt) and (d phi)/(dt)`, where i is the current in the circuit and `phi` is the flux of the megnetic field due to the long wire through the circuit.
(b) It is observed that at time t=T, the metal bar AB si at a distance of `2x_(0)` from the long wire and the resistance R carries a current `(i_1)`. Obtain an expression for the net charge that has flown through riesistance R form t=0 to t=T.
(c) THe bar is suddenly stopped at time T. THe current through resistance R is found to be `(i_1)/(4)` at time 2T. Find the value of `L/R` in terms of hte other given quantities.

Explanation

a. When bar `AB` slider away, the magneic force acts on free electrons along negative Y-axis, so electrons move from end `A` to end `B`, making end `A` at positive potential relative to end `B`. So current in the circuit will flow from `B` to `A` as shows.
Let `epsilon` be the induced emf According to kirchhoff's second law,
`-Ri - L(di)/(dt) + epsilon = 0 rArr epsilon = Ri + L (di)/(dt)`
But `epsilon = (d phi)/(dt)` (numerically)
`:. (d phi)/(dt) = Ri + L (di)/(dt)` (i)
This i sthe required relation.
b. Equation (i) can be expressed as
`d phi = Ridt + L di rArr d phi R dq + ldi`
Integrating, we get
`int_(0)^(T) d phi = R int_(0)^(q) dq + L int_(0)^(i_(1)) di`
`[phi]_(t = 0)^(T) = Rq + Li_(1)`
`:.` Charge flows from `t = 0` to `t = T` will be
`q = (1)/(R) [phi(T) - phi(0)] - (Li_(1))/(R)` (ii)
Change in flux during the displacement of bar from `x = x_(0)` to `x = 2 x_(0)` in time `T` is
`phi(T) - phi(0) = int_(x_0)^(2_(x_0)) BdA = int_(x_0)^(2_(x_0)) (mu_(0)I_(0))/(2 pi x) ldx`
`= (mu_(0)I_(0)l)/(2 pi) log_(e) 2` (iii)
`:.` From Eq.(ii), charge flows
`q = (mu_(0)I_(0)l)/(2 piR) log_(e) 2 - (Li_(1))/(R)` (iv)
c. The equation of decay of current in an `LR` circuit is given by
`i = i_(0)e^(-Rt//L)` (v)
Given `i = (i_(1))/(4), i_(0) = i_(1), t = 2T - T = T`
`:.` Substituting these values in Eq.(v), we get
`(i_(1))/(4) = i_(1) e^(-RT//L) rArr e^(-RT//L) = (1)/(4)`
or `(RL)/(L) = log_(e)4 :. (L)/(R) = (T)/(log_(e) 4)`