Two parallel vertical metallic rails `AB` and `CD` are separated by `1m`. They are connected at the two ends by resistances `R_1` and `R_2` as shown in the figure. A horizontal metallic bar `l` of mass `0.2 kg` slides without friction, vertically down the rails under the action of gravity. There is a uniform horizontal magnetic field of `0.6 T` perpendicular to the plane of the rails. It is observed that when the terminal velocity is attained, the powers dissipated in `R_1` and `R_2` are `0.76W` and `1.2W` respectively `(g=9.8m//s^2)`

The terminal velocity fo the bar L will be

Options

A. 5 m/s
B. 1m/s
C. 2 m/s
D. non these

Answer

B

Explanation

The rod wil acquire terminal velocity only when magneitc force `F_(m)` = Bl1 due to electromagnetic induction balanced its weight , i.e.,
Bll-mg. `" " i.e. l = (0.2 xx 9.8 )/(0.6xx1 ) = (9.8)/(3)` A
now if e is the emf induced in the rod ,
e`xx` l = P = `P_(1) P_(2)`
e = `(0.76 + 1.20)/(9.8//3) ` 0.6V
So, Now as this e is generated due to motion of rod with lerminal velcoity in the magnetic field . i.e,
e = `Bu_(r)`l `" " "so" u_(r) = (e)/(Bl) = (0.6 )/(0.6 xx 1 ) 1ms^(-1)`
`therefore` (B)