A conductor rod `AB` of mass `m` slides without friction over two long conducting rails separated by a distance (Fig) At the left end the raidls are interconnected by a resistance `R`. The system is located in a unifrom magnetic fileld perpendicular to the plane of the loop. At the moment `t = 0` the rod `AB` starts moving to the right with an initial velocity `v_(0)`. Neglecting the resistances of the rails and the rod `AB`, as wellas the self -indcuctance, find:
(a) the distance covered by the rod until it comes to a standsill,
(b) the amount of heat generated in the resitance `R` during this process.

Explanation

(a) The flux through the loop changes due to the movement of the rod `AB`. According to Lenz's law current should be anticlockwise in sese as we have assumed `B` is directed into the plane of the loop. The motion `e.m.f xi_("in") (t) = B l v`
and induced current `i_(i n) = (v B l)/(R)`
From Newton's law in projection from `F_(x) =n mw_(x)`
`-F_(emp) = m(v dv)/(dx)`
But `F_(amp) = i_(i n) l B = (v B^(2) l^(2))/(R)`
So, `- (v B^(2) l^(2))/(R) = m v (dv)/(dx)`

or, `int_(0)^(x) dx = -(mR)/(B^(2) l^(2)) int_(v_(0))^(0) dv` or, `x = (m R v_(0))/(B^(2) l^(2))`
(b) From equaction of energy conservation, `E_(f) - E_(i) +` Heat liberated `= A_(cell) + A_(ext)`
`[0 - (1)/(2)m v_(0)^(2)] +` Heat liberated `= 0 +0`
So, heat liberated `= (1)/(2)m v_(0)^(2)`