A copper rod of length l is rotated about one end perpendicular to the magnetic field B with constant angular velocity ω. The induced e.m.f. between the two ends is 

(1) 12Bωl2

(2) 34Bωl2

(3) Bωl2

(4) 2Bωl2

Concept Videos :-

#6 | Motional EMF (Induced Electric Field)
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#8 | Problem Set2: Motional EM
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Concept Questions :-

Motional emf

(1) If in time t. the rod turns by an angle θ, the area generated by the rotation of rod will be

=12l×lθ =12l2θ

So the flux linked with the area generated by the rotation of rod

ϕ=B12l2θcos0=12Bl2θ=12Bl2ωt

And so e=dϕdt=ddt12Bl2ωt=12Bl2ω

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