ODBAC is a fixed rectangular conductor of negligible resistance (CO is not connected) and OP is a conductor which rotates clockwise with angular velocity ω (figure). The entire system is in uniform magnetic field B whose direction is along the normal to the surface of the rectangular conductor ABDC. The conductor OP is in electric contact with ABDC. The rotating conductor has a resistance of λ per unit length. Find the current in the rotating conductor, as it rotates by 180°.

                

Hint: The cutting of the magnetic field results in induced emf.
Step 1:
Letusconsiderthepositionofrotatingconductorattimeinterval
t=0tot=π4ωorT/8;
therodOPwillmakecontactwiththesideBD.LetthelengthOQofthecontactatsometimet
suchthat0<t<π4ωor0<t<T8bex.ThefluxthroughtheareaODQis;
ϕ=B12QD×OD=B12ltanθ×l
=12Bl2tanθ,whereθ=ω
Step 2: ApplyingFaraday'slawofEMI,
Thus,themagnitudeoftheemfgeneratedisε=dt=12Bl2ωsec2ωt
ThecurrentisI=εRwhereRistheresistanceoftherodincontact.
Where,Rλ
R=λx=λlcosωt
I=12Bl2ωλlsec2ωtcosωt=Blω2λcosωt
Step 3: LetthelengthOQofthecontactatsometimetsuchthatπ4ω<t<3π4ωorT8<t<3T8bex.
TherodisincontactwiththesideAB.ThefluxthroughtheareaOQBDis:
ϕ=l2+12l2tanθB
where,θ=ωt
Thus,themagnitudeofemfgeneratedintheloopis:
ε=dt=12Bl2ωsec2ωttan2ωt
ThecurrentisI=εR=ελx=εsinωtλl=12Blωλsinωt
Similarlyfor3π4ω<t<πωor3T8<t<T2,therodwillbeintouchwithAC.
Step 4: ThefluxthroughOQABDisgivenby:
ϕ=2l2-l22tanωtB
Andthemagnitudeofemfgeneratedinloopisgivenby:
ε=dt=Rωl2sec2ωt2tan2ωt
l=εR=ελx=12Blωλsinωt
Thesearetherequiredexpressions.