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 1/8th of its value at the center of the coil, is
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a
R/3
b
R3
c
2R3
d
23R
answer is B.
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Detailed Solution
Baxis=μ04π×2πIR2R2+x23/2At center, Bcentre=μ04π×2πIRIn the given problem,Baxial=18Bcenter ⇒μ04π×2πIR2R2+x23/2=18μ04π×2πIROr R2+x23/2=8R3Solving, we get x=R3 .
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 1/8th of its value at the center of the coil, is