A rod of mass `m` and radius `R` rest on two parallel rails that are distance `l` apart and have a length `L`. The rod carries a current `I`( in the direction shown ) and rolls along the rails without slipping. A uniform magnetic field `B` is directed perpendicular to the rod and the rails. If it starts from rest, wat is the speed of the rod as it leaves the rails ?

Options

A. `sqrt((4BIlL)/(3m))`
B. `sqrt((3BIlL)/(3m))`
C. `sqrt((4BIlL)/(2m))`
D. `sqrt((BIlL)/(m))`

Answer

A

Explanation

The magnetic force acting on the rod is given by

`vec(F)=I(vec(I)xxvec(B))=Il(hat(k))xxB(hat(j))=(BIl)hat(i)`
From Work Energy Theorem, we have
`(K_("trans")+K_("rot"))_("initial")+DeltaE=(K_("trans")+K_("rot"))_("final")`
`implies" "0+0+Fscostheta=(1)/(2)m upsilon^(2)+(1)/(2)Iomega^(2)`, ,brgt where `" "I=(1)/(2)mR^(2)`
`implies(BIl)Lcos0^(@)=(1)/(2)m upsilon^(2)+(1)/(2)((1)/(2)mR^(2))((upsilon)/(R))^(2)`
`impliesBIlL=(3)/(4)mupsilon^(2)impliesupsilon=sqrt((4BIlL)/(3m))`