Two parallel vertical metallic bars `X X^(1)` and `Y Y^(1)`, of negligible resistance and separated by a length `'l'`, are as shown in Fig. The ends of the bars are joined by resistance `R_(1)` and `R_(2)`. A uniform magnetic field of induction `B` exsits in space normal to the plane of the bars. A horizontal metallic rod `PQ` of mass `m` starts falling vertically, making contact with the bars. It is oberved that in the steady state the powers dissipated in the resistance `R_(1)` and `R_(2)` and the terminal velocity attained by the rod `PQ`.

Explanation

Let `V_(0)` be the terminal velocity attained by the rod `PQ` (in the steady state). If `i_(1)` and `i_(2)` be the currents flowing through `R_(1)` and `R_(2)` in this state, then current following through the rod `PQ` is `i = i_(1)+i_(2)` (see the circuit diagram) as shown in Fig.

`:.` Applying Kirchoff's loop rule, yeilds.
`i_(1)R_(1) = BV_(0)l` and `i_(2)R_(2) = BV_(0)l`
`:. i_(1) + i_(2) = BV_(0)l((1)/(R_(1))+(1)/(R_(2)))` .....(i)
Given that, `P_(1) = i_(1)^(2)R_(1) = (B^(2)V_(0)^(2)l^(2))/(R_(1))` .......(i)
and `P_(2) = i_(2)^(2)R_(2) = (B^(2)V_(0)^(2)l^(2))/(R_(2))` ....(iii)
Also in the steady state, the acceleration of `PQ = 0`
`rArr mg = B(i_(1)+i_(2))l`
(or) `mg = B^(2)l^(2)V_(0)((1)/(R_(1))+(1)/(R_(2))) = P_(1)+P_(2)`
[From equation (ii) and (iii)]
`:.` The terminal velocity is `V_(0) = (P_(1)+P_(2))/(mg)`
Substituting for `V_(0)` in equation (ii),
`P_(1) = (B^(2)l^(2))/(R_(1))((P_(1)+P_(2))/(mg))^(2) rArr R_(1) = [(Bl(P_(1)+P_(2)))/(mg)]^(2)xx(1)/(P_(1))`
Similarly from equation (iii)
`R_(2) = [(Bl(P_(1)+P_(2)))/(mg)]^(2)xx(1)/(P_(2))`