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. A horizontal metallic bar PQ of mass 0.2 kg slides without friction, vertically down the rails under the action of gravity. There is uniform horizontal magnetic field of 0.6T perpendicular to plane of the rails. It is observed that when the terminal velocity attained, the power dissipated in $R_{1}$ and $R_{2}$ are 0.76 W and 1.2 W respectively. Find the terminal velocity of bar in m/s. $(g = 9.8 m / s^{2})$

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Detailed Solution

When bar attains terminals velocity $B I L = m g$

$I = \frac{0.2 \times 9.8}{0.6 \times 1} = \frac{9.8}{3},$ e is emf induced in rod and $e I = P_{1} + P_{2}$

$e = \frac{0.76 + 1.2}{9.8 / 3} = \frac{1.96 \times 3}{9.8} = 0.6 V$ and $e = B v_{T} l$

$v_{T} = \frac{e}{B} = 1 m / s$

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