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An Intiative by Sri Chaitanya
a
R
b
R2
c
R2
d
R1-12
answer is C.
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Detailed Solution
ϕA=μ0iπR22π(R2+x2)3/2πr2⇒EA=-dϕdt=-μ0iπ2R2r2(-3/2)(R2+x2)-5/22xEA is maximum when dEAdx=0⇒ddxx(R2+x2)5/2=0or (R2+x2)5/2-5x2(R2+x2)3/22x=0or, R2+x2-5x2=0or, x=R2