Two identical long conducting wires AOB and COD are placed at right angle to each other, with one above other such that O is their common point for the two. The wires carry I₁ and I₂ currents, respectively. Point P is lying at distance d from O along a direction perpendicular to the plane containing the wire. The magnetic field at the point P will be

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$\frac{μ_{0}}{2 πd} (\frac{I_{1}}{I_{2}})$

$\frac{μ_{0}}{2 πd} (I_{1}^{2} + I_{2}^{2})$

$\frac{μ_{0}}{2 πd} (I_{1} + I_{2})$

$\frac{μ_{0}}{2 πd} {(I_{1}^{2} + I_{2}^{2})}^{1 / 2}$

answer is D.

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Detailed Solution

The magnetic field at the point P, at a perpendicular distance d from O in a direction perpendicular to the plane containing A, B, C and D due to currents through AOB and COD are perpendicular to each other. Hence,

$B = {(B_{1}^{2} + B_{2}^{2})}^{1 / 2}$

$= {[{(\frac{μ_{0}}{2 π} \frac{I_{1}}{d})}^{2} + {(\frac{μ_{0}}{2 π} \frac{I_{2}}{d})}^{2}]}^{1 / 2}$

$= \frac{μ_{0}}{2 πd} {(I_{1}^{2} + I_{2}^{2})}^{1 / 2}$

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