Consider the situation shown in the figure o fthe previous problem. Suppose the wire connecting O and C has zero resistance but circular loop has a resistance R uniformly distributed along its length. The rod OA is made ot rotate with a uniform angular speed `omega` as shown in the figure. Find the current in the rod when `ltAOC = 90^@`.

Explanation

The two resistances `(r/4) and (3r)/(4) are in parallel.`
Equivalent resisatance R'=((r/4)xx(3r/4)/®=(3r)/(16)`
`e=Bvl= Bxx(a/2) omega xxa`
`=(Ba^2 omega)/2`
`e/(R')=(Ba^2 omega)/(2R')=(Ba^2 omega)/(2xx 3r//16)`
`=(Ba^2 omega 16)/(2xx3r)=(8Ba^2 omega)/r`.