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The loop shown in figure in carrying a current I the magnetic induction at the common centre O of the semi circles is B. Now the wire of the loop is bent to form a circular ring which is carrying the same current I. Then magnetic induction at the centre of the ring will be

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a

8πB/3(3π+2)

b

4πB/(π+2)

c

8B/(π+4)

d

2B/(π+8)

answer is A.

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Detailed Solution

B=μ0I4R+μ0I4(R2)=3μ0I4RIf radius of the circular ring be R1, then 2πR1=πR+π(R2)+R⇒R1=3πR2×2π+R2π=(3π+24π)R∴ B1=μ0I2R1=μ0IR(3π+2)/2π=μ0I×2πR(3π+2)⇒B1=2π(3π+2)×4B3=8πB3(3π+2)

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