There are two infinitely long straight current carrying conductors and they are held at right Angles to each other so that their common ends meet at the origin as shown in the figure given Below. The ratio of current in both conductors is 1:1. The magnetic field at point P is

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$\frac{μ_{0} I}{4 π x y} [\sqrt{x^{2} + y^{2}} + (x + y)]$

$\frac{μ_{0} I x y}{4 π} [\sqrt{x^{2} + y^{2}} + (x + y)]$

$\frac{μ_{0} I x y}{4 π} [\sqrt{x^{2} + y^{2}} - (x + y)]$

$\frac{μ_{0} I}{4 π x y} [\sqrt{x^{2} + y^{2}} - (x + y)]$

answer is A.

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Detailed Solution

$B_{1} = \frac{μ_{0} I}{4 π y} [\sin 90 + \sin θ_{1}] = \frac{μ_{0} I}{4 π y} (1 + \frac{x}{\sqrt{x^{2} + y^{2}}}) .... (i)$

$B_{2} = \frac{μ_{0} I}{4 π x} (\sin 90 + \sin θ_{2}) = \frac{μ_{0}}{4 π} \frac{I}{y} (1 + \frac{y}{\sqrt{x^{2} + y^{2}}}) ..... (i i)$

$B_{N e t} = B_{1} + B_{2}$

$B = \frac{μ_{0} I}{4 π} [\frac{1}{y} + \frac{x}{y \sqrt{x^{2} + y^{2}}} + \frac{1}{x} + \frac{y}{x \sqrt{x^{2} + y^{2}}}] \Rightarrow B = \frac{μ_{0} I}{4 π} [\frac{x + y}{x y} + \frac{x^{2} + y^{2}}{x y \sqrt{x^{2} + y^{2}}}]$

$= \frac{μ_{0} I}{4 π} [\frac{x + y}{x y} + \frac{\sqrt{x^{2} + y^{2}}}{x y}] \Rightarrow B = \frac{μ_{0} I}{4 π x y} [\sqrt{x^{2} + y^{2}} + (x + y)]$