1.

A rod of 5m length is moving perpendicular to uniform magnetic field of intensity 2 xx 10^(-4)"Wb/m"^2. If the acceleration of rod is 2ms^(-2), th rate of increase of the induced emf is____

Answer»

`20xx10^(-4) "V/sec"^2`
`20xx10^(-4)V`
`20xx10^(-4)` Vs
`20xx10^(-4)` V/s

Solution :Magnitude of induced emf E=Bvl
Taking derivation w.r.t. t,`"dE"/"dt"=d/"dt"` (Bvl)
B and L are constant `therefore (dE)/(dt)=B(DV)/(dt)l`
`therefore (dE)/(dt)=Bal "" [because (dv)/(dt)=a]`
`=2xx10^(-4)xx2xx5`
`therefore (dE)/(dt)=20xx10^(-4)` V/s


Discussion

No Comment Found

Related InterviewSolutions