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

Options

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))`

Answer

A

Explanation

(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))`