A conducting wire of length `l` and mass `m` is placed on two inclined rails as shown in figure. A current `I` is flowig in the wire in the direction shown When no magnetic field is present in the region, the wire is just on the verge of sliding. When a vertically upward magnetic field is switched on, the wire starts moving up the incline. The distance travelled by the wire as a function of time `t` will be

Options

A. `(1)/(2)[(IBl)/(m)-2g]t^(2)`
B. `(1)/(2)[(IBl)/(m)xx(1)/(cos theta)-2g sin theta]t^(2)`
C. `(1)/(2)[(IBl)/(m)-2g sin theta]t^(2)`
D. `(1)/(2)[(IBl)/(m)(cos 2 theta)/(cos theta)-2g sin theta]t^(2)`

Answer

D

Explanation

The front view of the arrangment is shown in figure

From initial condition, `mg sin theta = mu g cos theta`
`rArr mu = tan theta`
`ma = IBl cos theta - mg sin theta-mu N`
`N = mg cos theta + IBl sin theta`
`rArr a = (IBl)/(m) cos alpha q- 2g sin theta - (IBl)/(m) (sin^(2) theta)/(cos theta)`
`= (IBl)/(m) (cos 2 theta)/(cos 2 theta)-2 g sin theta`
Now,
`s = (1)/(2)a t^(2) = (1)/(2)[(IBl)/(m)(cos 2 theta)/(cos theta)-2 g sin theta]t^2`