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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_2A 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_2respectively. Find an expression for R_1,R_2 and the terminal velocity attained by the rod PQ. |
Answer» Solution : 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` [FromEq. (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)` |
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