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A circular current carrying coil has a radius R. The distance from the center of the coil on the axis where the magnetic induction will be (1/8)th of its value at the center of the coil,is

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a

R/3

b

R3

c

2R3

d

(23)R

answer is B.

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Detailed Solution

Baxis=μ04π×2πIR2R2+x23/2 At centre, Bcentre =μ04π×2πIR In the given problem, μ04π×2πIR2R2+x23/2=18μ04π×2πIR or R2+x23/2=8R3 Solving, we get x=R3

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