A copper rod of mass `m` slides under gravity on two smooth parallel rails `l` distance apart set at an angle `theta` to the horizontal. At the bottom, the rails are joined by a resistance `R`.

There is a uniform magnetic field perpendicular to the plane of the rails. the terminal valocity of the rod is

Options

A. `(mg R tan theta)/(B^(2)l^(2))`
B. `(mg R cot theta)/(B^(2)l^(2))`
C. `(mg R sin theta)/(B^(2)l^(2))`
D. `(mg R cos theta)/(B^(2)l^(2))`

Answer

C

Explanation

Terminal velocity of the rod is attained when magnetic force on the rod (Bil) balances the components of weight of the rod `(mg sin theta)`, as in figure.

So, Bil `=mg sin theta`
`B ((e )/(R ))l = mg sin theta`
`(B l e)/(R )= mg sin theta`
`(B l (Bl v_(T)))/(R )= mg sin theta`
`V_(T)= (mg sin theta R)/(B^(2)l^(2))`