Two parallel resistanceless rails are connected by an inductor of inductance `L` at one end as shows in Fig. A magnetic field `B` exists in the space which is perendicular to the plane of the rails. Now a conductor of length `l` and mass `m` is placed transverse on the rail and given an inpulse `J` toward the rightward direction. Then choose the correct option (S). A. Velocity of the conductor is half of the initial velocity after a displaacement of the conductor `d = sqrt((3J^(2)L)/(4B^(2)l^(2)m))`B. Current flowing through the inductor at the instant when velocity of the conductor is half of the initial velocity is `i = sqrt((3J^(2))/(4Lm))`C. Velocity of the conductor is half of the initial velocity after a displacement of the conductor `d = sqrt((3J^(2)L)/(B^(2)l^(2)m))`D. Current flowing through the inductor at the instant when velocity of the conductor is half of the initial velocity is `i = sqrt((3j^(2))/(mL))`

Two parallel resistanceless rails are connected by an inductor of indu

1.	Two parallel resistanceless rails are connected by an inductor of inductance `L` at one end as shows in Fig. A magnetic field `B` exists in the space which is perendicular to the plane of the rails. Now a conductor of length `l` and mass `m` is placed transverse on the rail and given an inpulse `J` toward the rightward direction. Then choose the correct option (S). A. Velocity of the conductor is half of the initial velocity after a displaacement of the conductor `d = sqrt((3J^(2)L)/(4B^(2)l^(2)m))`B. Current flowing through the inductor at the instant when velocity of the conductor is half of the initial velocity is `i = sqrt((3J^(2))/(4Lm))`C. Velocity of the conductor is half of the initial velocity after a displacement of the conductor `d = sqrt((3J^(2)L)/(B^(2)l^(2)m))`D. Current flowing through the inductor at the instant when velocity of the conductor is half of the initial velocity is `i = sqrt((3j^(2))/(mL))`
Answer» Correct Answer - A::B `L(di)/(dt) = B nu l rArr int di = (Bl)/(L) int nu dt rArr I = (Bl)/(L) x` (i) `F = ma rArr - iBl = m nu (d nu)/(dx)` `rArr -(B^(2)l^(2)x)/(L) = mnu (d nu)/(dx) rArr -(B^(2)l^(2))/(L) underset(0) overset(d) int x dx = underset(nu _(0)) overset(nu_(0)//2)int nu dnu` `rArr -(B^(2)l^(2)d^(2))/(L) = (- 3 nu_(0)^(2))/(8), nu_(0) = (J)/(m)` `rArr d = sqrt((3J^(2)L)/(4B^(2)l^(2)m))` Put `x = d` in (i), `i = (Bl)/(L) sqrt((3J^(2)L)/(4B^(2)l^(2)m)) = sqrt((3J^(2))/(4Lm))`

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