A rod of mass 0.720 kg and radius 6 cm rests on two parallel rails that are `d=12 cm` apart and length of rails is `L=49 cm.` The rod carried a current of `I=48 A` Fig. and rolls along the rails without slipping. A uniform magnetic field of magnetude 0.240 T is directed perpendicular to the rod and the rails. If it starts from rest, what is the speed of the rod as it leaves the rails?

Explanation

`F=idB=48xx12/100xx0.240=1.3824N`

`F-f=ma and fR=Ialpha implies fR =Ia/R`
`implies F-(Ia)/(R^2)=ma`
`implies a=F/(m+I/R^2)=F/(m+1/2 (mR^2)/(R^2))=(2F)/(3m)`
`=(2xx1.3824)/(3xx0.720)=1.28 m//s^2`
Now, `v^2=u^2+2as`
`implies v=sqrt(2xx1.28xx0.49)=1.12 m//s`
Alternatively:
`W=DeltaKE implies FL=1/2mv^2+1/2Iomega^2`
`implies IdBL=1/2mv^2+1/2(1/2mr^2)omega^2`
`implies IdBL=3/4mv^2 implies v=sqrt(4/3 (IdBL)/m)=1.12ms^-1`