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Q.

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
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