ϕ2=B1×A2=2(R12+x2)23μ0IR12×πR22
= 2[(0.2)2+(0.15)2]23μ0(2)(20×10−2)2×π(0.3×10−2)
On solving
=9.216×10−11
≈9.2×10−11 weber
JEE Main 2013 — Physics Electromagnetism
A circular loop of radius 0.3cm lies parallel to a much bigger circular loop of radius 20cm. The centre of the small loop is on the axis of the bigger loop. The distance between their centres is15 cm. If a current of 2.0A flows through the smaller loop, then the flux linked with a bigger loop is:
Held on 7 Apr 2013 · Verified 6 Jul 2026.
3.3×10−11weber
6.6×10−9weber
9.1×10−11weber
6×10−11weber
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