Two identical long conducting wires AOB and COD are placed at right...

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

{I}_{1}

and

{I}_{2}

currents, respectively. Point ‘

I

’ is lying at distance ‘

d

’ from ‘

O

’ along a direction perpendicular to the plane containing the wires. The magnetic field at the point ‘

P

’ will be: Two identical long conducting wires AOB and COD are placed at right angle to each other, with one above other such that

Options

A $\dfrac{{\mu}_{0}}{2\pid}\left(\dfrac{{I}_{1}}{{I}_{2}}\right)$
B $\dfrac{{\mu}_{0}}{2\pid}\left({I}_{1} + {I}_{2}\right)$
C $\dfrac{{\mu}_{0}}{2\pid}\left({I}_{1}^{2} - {I}_{2}^{2}\right)$
D $\dfrac{{\mu}_{0}}{2\pid}{\left({I}_{1}^{2} + {I}_{2}^{2}\right)}^{1 / 2}$

Correct answer

D. $\dfrac{{\mu}_{0}}{2\pid}{\left({I}_{1}^{2} + {I}_{2}^{2}\right)}^{1 / 2}$

Step-by-step solution

Two identical long conducting wires AOB and COD are placed at right angle to each other, with one above other such that
$\overset{\to}{B}$ due to wire \left(1\right){\overset{\to}{B}}_{1} = \left(\dfrac{{\mu}_{0}{i}_{1}}{2\pid}\right)\overset{^}{i} $<br />$ \overset{\to}{B} $due to wire \left(2\right){\overset{\to}{B}}_{2} = \left(\dfrac{{\mu}_{0}{i}_{2}}{2\pid}\right)\left(- \overset{^}{j}\right)$
\left|{B}_{net}\right| = \dfrac{{\mu}_{0}}{2\pid}\sqrt{{i}_{1}^{2} + {i}_{2}^{2}}

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