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Question
A 20cm long conducting rod is set into pure translation with a uniform velocity of 10cms−1 perpendicular to its length. A uniform magnetic field of magnitude 0.10T exists in a direction perpendicular to the plane motion. (a) Find the average magnetic force on the free electrons of the rod. (b) For what electric field inside the rod, the electric force on a free electron will balance the magnetic force? How is this electric field created? (c) Find the motional emf between the ends of the rod.
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Solution
l=20cm=0.2m;v=10cm/s=0.1m/s;B=0.10T
(a) F=qvB=1.6×10−19×1×10−1×1×10−1=1.6×10−21N
(b) qE=qvB⇒E=1×10−1×1×10−1=1×10−2V/m
this is created due to the induced emf
(c) Motional emf=Bvl
=0.1×0.1×0.2=21×10−3V
(a) F=qvB=1.6×10−19×1×10−1×1×10−1=1.6×10−21N
(b) qE=qvB⇒E=1×10−1×1×10−1=1×10−2V/m
this is created due to the induced emf
(c) Motional emf=Bvl
=0.1×0.1×0.2=21×10−3V
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