QJEEAdv14Ph7571: JEE Advanced 2014 Physics - Class 12 JEEAdv14Ph7571 Solution: When d a but wires are not touching the loop, it is found that…

B_R = B due to ring

B₁ = B due to wire-1

B₂ = B due to wire-2

In magnitudes

${B_1} = {B_2} = {{{\mu _0}I} \over {2\pi r}}$

Resultant of B₁ and B₂

$= 2{B_1}\cos \theta = 2\left( {{{{\mu _0}I} \over {2\pi r}}} \right)\left( {{h \over r}} \right)$

$= {{{\mu _0}Ih} \over {\pi {r^2}}}$

${B_R} = {{{\mu _0}I{R^2}} \over {2{{({R^2} + {x^2})}^{3/2}}}}$

$= {{2{\mu _0}I\pi {a^2}} \over {4\pi {r^3}}}$

As, R = a, x = h and a² + h² = r²

For zero magnetic field at P,

${{{\mu _0}Ih} \over {\pi {r^2}}} = {{2{\mu _0}I\pi {a^2}} \over {4\pi {r^3}}}$

$\Rightarrow \pi {a^2} = 2rh \Rightarrow h \approx 1.2a$