A straight wire of length L is bent into a semicircle. It is moved in a uniform magnetic field with speed v with diameter perpendicular to the field. The induced emf between the ends of the wire is
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a
BLv
b
2BLv
c
2πBLv
d
2BvLπ
answer is D.
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Detailed Solution
Induced emf, e=Bvle where le is the effective length between the ends of the bent rod . ⇒e=Bv(2R)=2BvLπHere le =2R = 2(Lπ) Therefore e = 2vLBπ