We know, the magnetic field on the axis of a current carrying circular ring is given by
B=4πμ0(R2+x2)3/22NIA
∴B2B1=18=[R2+(0.05)2R2+(0.2)2]3/2
4[R2+(0.05)2]=[R2+(0.2)2]
4R2−R2=(0.2)2−4×(0.05)2
4R2−R2=(0.2)2−(0.1)2
3R2=0.3×0.1
R2=(0.1)2⇒R=0.1
JEE Main 2021 — Physics Electromagnetism
Magnetic fields at two points on the axis of a circular coil at a distance of 0.05m and 0.2m from the centre are in the ratio 8:1. The radius of coil is _______ .
Held on 25 Feb 2021 · Verified 6 Jul 2026.
0.15m
0.2m
0.1m
1.0m
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