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:
Let us consider the position of rotating conductor at time interval
t=0 to t=π4ωor T/8;
the rod OP will make contact with the side BD. Let the length OQ of the contact at sometime t
such that 0<t<π4ωor 0<t<T8 be x. The flux through the area ODQ is;
ϕ=B12QD×OD=B12ltanθ×l
  =12Bl2 tanθ, where θ=ω
Step 2: Applying Faraday's law of EMI,
Thus, the magnitude of the emf generated is ε=dt=12Bl2ωsec2ωt
The current is I=εR where R is the resistance of the rod in contact.
Where, Rλ
R=λx=λlcosωt
  I=12Bl2ωλlsec2ωtcosωt=Blω2λcosωt
Step 3: Let the length OQ of the contact at some time t such that π4ω<t<3π4ωorT8<t<3T8be x.
The rod is in contact with the side AB. The flux through the area OQBD is:
ϕ=l2+12l2tanθB
where,        θ=ωt
Thus, the magnitude of emf generated in the loop is:
ε=dt=12Bl2ωsec2ωttan2ωt
The current is I=εR=ελx=εsinωtλl=12Blωλsinωt
Similarly for 3π4ω<t<πωor3T8<t<T2, the rod will be in touch with AC.
Step 4: The flux through OQABD is given by:
ϕ=2l2-l22 tan ωtB
And the magnitude of emf generated in loop is given by:
ε=dt=Rωl2sec2ωt2 tan2ωt
l=εR=ελx=12Blωλsinωt
These are the required expressions.