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a conducting wire os mass `m` slides down two smooth conducting bars, set at an angle `theta` to the horizontal as shown in . The separatin between the bars is `l`. The system is located in the magnetic field `B`, perpendicular to the plane of the sliding wire and bars. The constant velocity of the wire is A. (a) `(mgR sin theta)/(B^(2)l^(2))`B. (b) `(mgR sin theta)/(Bl^(3))`C. ( c) `(mgR theta)/(B^(2)l^(5))`D. ( d) `(mgR sin theta)/(Bl^(4))` |
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Answer» Correct Answer - A (a) Component of weight along the inclined plane `= mg sin theta` Again, `F = Bil = B(Nlv)/(R )l = (B^(2)l^(2)v)/(R )` Now, `(B^(2)l^(2)v)/(R ) = mg sin theta` or `v = (mg R sin theta)/(B^(2)l^(2))` |
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