Two coplanar circular coils of equal radius carrying currents $i_{1}, i_{2}$ in opposite directions are at a large distance ‘d’. The distance from the first coil where the resultant magnetic induction is zero is

see full answer

21% of IItians & 23% of AIIMS delhi doctors are from Sri Chaitanya institute !!

An Intiative by Sri Chaitanya

$\frac{d}{1 + {(\frac{i_{2}}{i_{1}})}^{2}}$

$\frac{d}{1 + \sqrt{\frac{i_{2}}{i_{1}}}}$

$\frac{d}{1 + \frac{i_{2}}{i_{1}}}$

$\frac{d}{1 + {(\frac{i_{2}}{i_{1}})}^{\frac{1}{3}}}$

answer is C.

(Unlock A.I Detailed Solution for FREE)

Ready to Test Your Skills?

Check your Performance Today with our Free Mock Test used by Toppers!

Take Free Test

Detailed Solution

Considering two loops

$Magnetic field on axis of coil = \frac{μ_{0} i a^{2}}{2 {(a^{2} + x^{2})}^{3 / 2}}$

here $d > > x and a > > x$

$B = \frac{μ_{0} {ia}^{2}}{2 x^{3}}$

At point P

$B_{1} = \frac{μ_{0} i_{1} a^{2}}{2 x^{3}}, B_{2} = \frac{μ_{0} i_{2} a^{2}}{2 (d - x)^{3}}$

$B_{1} = B_{2}$

$\Rightarrow \frac{μ_{0} i_{1} a^{2}}{2 x^{3}} = \frac{μ_{0} i_{2} a^{2}}{2 (d - x)^{3}} \Rightarrow \frac{i_{2}}{i_{1}} = {(\frac{d - x}{x})}^{3}$

$\Rightarrow \frac{d - x}{x} = {(\frac{i_{2}}{i_{1}})}^{1 / 3}$

$x = \frac{d}{1 + {(\frac{i_{2}}{i_{1}})}^{1 / 3}}$

Watch 3-min video & get full concept clarity