A current carrying conductor AB is bent in a form of a semi circle of radius r and it is placed in a uniform magnetic field of induction B as shown in figure. Then force experienced by the conductor is
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a
Zero
b
2irB
c
irB
d
2πirB
answer is A.
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Detailed Solution
F→=ile→×B→. Here equivalent length le is parallel to B→.Hence F=ileBsinθo=0