Two parallel vertical metallic bars XX.and YY., 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 exists 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 observed that in the steady state the powers dissipated in the resistance `R_(1)` and `R_(2)` are `P_(1)` and `P_(2)` respectively. Find an expression for `R_(1), 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 flowing through the rod PQ is `i= i_(1) + i_(2)` (see the circuit diagram) showin in fig.

`therefore` Applying Kirchhoff.s loop rule, yields `i_(1) R_(1)= Bv_(0)l and i_(2)R_(2)= Bv_(0)l`
`therefore 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))` ...(ii) 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. `mg= B (i_(1) + i_(2))l`
`mg= B^(2)l^(2)v_(0) ((1)/(R_(1)) + (1)/(R_(2)))` ...(iv)
Multiplying both sides by `v_(0)`
`mgv_(0)= B^(2) l^(2) v_(0)^(2) ((1)/(R_(1)) + (1)/(R_(2)))= P_(1) + P_(2)`
[From Eq. (ii) and (iii)]
`therefore` The terminal velocity is `v_(0)= (P_(1) + P_(2))/(mg)`
Substituting for `v_(0)` in Eq. (ii),
`P_(1)= (B^(2)l^(2))/(R_(1)) ((P_(1) + P_(2))/(mg))^(2), R_(1)= [(Bl(P_(1)+ P_(2)))/(mg)]^(2) xx (1)/(P_(1))`
Similarly from Eq. (iii)
`R_(2)= [(Bl(P_(1) + P_(2)))/(mg)]^(2) xx (1)/(P_(2))`