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.

Answer

A, B, C, D

Explanation

As the metal bar AB moves towards the right the magnetic flux in the loop. ABCD increases in the downward direction. By Lenz's law, to oppose this current will flow inanticlock wise direction as shown in figure.
Applying Kirchhoff's loop law in ABCD, we get
`(d phi)/(dt)=iR+L(di)/(dt(` ...(i)(b) Let AB beat a distance x fro the long striaght wire at any instant of time t during its mition, The magnetic field at that instant at AB due to lang striaght current carrying wire is
`B=(mu_(0)I_(0))/2 pi x)`
The change in flux through ABCD in time dt is
`(d phi)=B(dA)=Bldx`
Therefore, the total flux change when metal bar moves from a distance `(x_0) to 2(x_0)` is
`(Delta phi)=int _(x_0)^(2x_(0)) Bldx=l int_(x_0)^(2x_(0)) (mu_(0)I_(0))/(2 pi) dx=(mu_(0)I_(0)l)/(2 pi)[log_(e)x]_(x_0)^(2x_(0))`
`=(mu_(0)I_(0)l)/(2 pi) log_(e)2` ....(ii) THe charge flowing through resistance R in time T is
`q=int_(0)^(T) idt int_(0)^(T) [E_(induced)-L(di)/(dt)]dt (from eq.(i))
`=1/R int_(0)^(T) E_(induced) dt -L/R int _(0)^(l_1) d i=1/R (Delta phi)-L/R(i_1)`
q=1/R[(mu_(0)I_(0)l)/(2 pi)log_(e)2]-L/R(i_1) from eq(ii)`
(c) When the metal bar AB is stopped, the rate of change of magnetic flu through ABCD bexomes zero. from (i),
`iR=-L(di)/(dt)`
`int_(T)^(2T) dt=L/R int_(i_1)^(i_(1)//4) (di)/i`
T==-L/R log_(e) (i_(1)//4)/(i_1) implies L/R=T/(2 log_(e)2)`.