Two parallel vertical metellic rails `AB` and `CD` are separated by `1m`. They are connecting at two ends by resistances `R_(1)` and `R_(2)` as shown in Fig. 3.96. A horizontal metallic bar `L` mas `0.2 kg` slides without friction vertically down the rails under the action of gravity. There is a uniform horozontal magnetic field of `0.6 T` perpendicular to the plane of the rails. It is observed that when the terminal velocity is attained, thwe power dissipated in `R_(1) and R_(2)` are `0.76 and 1.2W`, respectively. Find the terminal velocity of the bar `L` and the values of `R_(1) and R_(2)`.

Explanation

Let magnetic field `B = 0.6 T` be inward as shown Fig. 3.97. When the rod moves down due to gravity, emf is induced in the rod as shown. Due to this, current `I` flows in the rod which results in force `F` on the rod in upward direction. Let at any instant of time, velocity of rod be `v`.
Then `e = Bvl and I = (e)/(R_(eq))`,
where `(1)/(R_(eq)) = (1)/(R_(1)) + (1)/(R_(2))`
`F = IlB = (e)/(R_(eq))lB = (B^(2)l^(2)v)/(R_(eq))`
Initially, `F` is less than mg. But as the velocity increases, `F` also increases. At a certain time, `F` will be equal to mg. At that time net force on the rod will become zero and velocity of rod will velocity. Let it be `v_(t)`. So putting `F = mg`
`rarr` `(B^(2)l^(2)v_(t))/(R_(eq)) = mg` ltbr. But at this time, power disspated in resistances is
`P_(1) + P_(2) = 0.76 + 1.2 = 1.96 W`
So we have `(e^(2))/(R_(eq)) = P_(1) + P_(2)` `rarr` `(B^(2)l^(2)v_(t)^(2))/(R_(eq)) = 1.96` ltbr. From Eqs. (i) and (ii) `v_(t) = (1.96)/(mg) = (1.96)/(0.2 xx 9.8) = 1 m s^(-1)`
From Eqs. (i) and `R_(eq) = (B^(2)l^(2)v_(t))/(mg) = ((0.6)^(2) xx 1^(2) xx 1)/(0.2 xx 9.8) = (9)/(49)Omega`
`P_(1)/(P_(2)) = (e^(2)//R_(1))/(e^(2)//R_(2))` `rarr` `(0.76)/(1.20) = (R_(2))/(R_(1))` `rarr` `(19)/(30) = (R_(2))/(R_(1))`
`(1)/(R_(eq)) = (1)/(R_(1)) + (1)/(R_(2))` `rarr` `(49)/(9) = (1)/(R_(1)) + (30)/(19R_(1))`
`rarr` `R_(1) = (9)/(19)Omega` and `R_(2) = (19)/(30)R_(1) = (19)/(30) xx (9)/(19) = 0.3 Omega`