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, and R, in this state, then current flowing through the rod PQ is i= `i_1+i_2` (see the circuit diagram) as shown in Fig.

` therefore ` Applying Kirchoff.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_i = 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.
` rArr mg= B (i_1 + i_2 )l`
or `mg = B^2 l^2 v_0 ((1)/(R_1) + (1)/(R_2))` ...(iv)
Multiplying both sides by `v_0` , we get
`mg v_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 rArr 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)`