A conducting frame is placed in a horizontal plane as shown in the figure. The two sides are parallel and separated by a distance of 0.25 m. A massless conducting rod can slide without friction on the frame. The total resistance of the circuit is 40 Omega. The rod is connected to a 0.2 kg mass by a massless cord which passes over a massless and frictionless pulley. A uniform magnetic field of 2T points vertically upward. The voltage of the bettery is 100 V. Find the constant velocity in m//s with which the rod and mass eventually move.

A conducting frame is placed in a horizontal plane as shown in the fig

1.	A conducting frame is placed in a horizontal plane as shown in the figure. The two sides are parallel and separated by a distance of 0.25 m. A massless conducting rod can slide without friction on the frame. The total resistance of the circuit is 40 Omega. The rod is connected to a 0.2 kg mass by a massless cord which passes over a massless and frictionless pulley. A uniform magnetic field of 2T points vertically upward. The voltage of the bettery is 100 V. Find the constant velocity in m//s with which the rod and mass eventually move.
Answer» Solution :When the ROD moves, with CONSTANT velocity, net force on the bar is zero :.W = GRAVITATIONAL force `=mg=i//B` [i=INDUCED current in the circuit] `:. I = (0.2xx10)/(2xx0.25)=4A` To produce 4A current in the bar, induced emf varepsilon in the circuit is `(100+varepsilon)/40 = 4 varepsilon=60V` We know, `varepsilon=BlV RIGHTARROW V=varepsilon/(Bl)=60/(2xx0.25)=120m//s`

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