A metallic rod of mass m length R is pivoted at O. It is free to rotate about a horizontal axis passing through O . An inward magnetic field is applied as shown in the figure. Find the ratio of maximum electric field inside the metallic rod to the applied magnetic field induction .

Explanation

The induced emf between O and P
`epsilon =-(dphi)/(dt) = -(d(BA))/(dt) = -B""(dA)/(dt)`
`epsilon = -B""(d)/(dt) [(r^(2))/(2) theta]= -(Br^(2)omega)/(2) (because omega=(d theta)/(dt))`
The potential gradient `(d epsilon )/(dr) =((-Br^(2)omega)/(2)) (d)/(dr) (r^(2)) = -B omega r`
`implies ` The corresponding electric field E = `B omegar`
`implies E_("max")= B omega` R where `omega` = maximum angular speed of the rod that is possible when it becomes vertical . Conservation of meachanical energy between horizontal and vertical yield ,
`Delta PE_(gr)+DeltaKE_(r)=0`
`-mg""(R)/(2)+(1)/(2)I_(0)omega^(2)=0`
`implies omega= sqrt((3g)/(R)) R `
`implies (E_("max"))/(B) = sqrt(3gR)`