A wire of length `l`, mass `m` and resistance `R` slides without any friction down the parallel conducting rails of negligible resistance . The rails are connected to each other at the bottom by a resistanceless rail parallel to the wire so that the wire and the rails form a closed rectangualr conducting loop. The plane of the rails makes an angle `theta` with the horizontal and a uniform vertical magnetic field of the inducetion `B` exists throughout the rregion. Find the steady state velocity of the wire.

Options

A. (a) `(mg)/(R )(sin theta)/(B^(2)l^(2)cos^(2)theta)`
B. (b) `(mg)/(R )(sin^(2) theta)/(B^(2)l^(2)cos^(2)theta)`
C. ( c) `(mgR sin theta)/(B^(2)l^(2)cos^(2)theta)`
D. ( d) `mgR (sin^(2) theta)/(B^(2)l^(2)costheta)`

Answer

C

Explanation

( c) `Blv cos theta = mg sin theta`
Here induced emf across slider is `(Bcos theta)lv`
`:.` Induced current `I = (Blv cos theta)/(R )`

From equation (i)
` Blcos theta(Blvcostheta)/(R ) = mg sin theta`
`:.` `v = (mg R sin theta)/(B^(2)l^(2)cos^(2) theta)`