Two infinitely long straight parallel wires are 5 m apart, perpendicular to the plane of paper. One of the wires, as it passes perpendicular to the plane of paper, intersects it at A and carries current i in the downward direction. The other wire intersects the plane A of paper at point B and carries current i in the outward direction O in the plane of paper as shown in figure. With x and y axes as shown, magnetic induction at O in the component form can be expressed as

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$\vec{B} = \frac{μ_{0} i}{4 π} [- 3 \hat{i} + \frac{7 \hat{j}}{18}]$

$\vec{B} = \frac{μ_{0} i}{2 π} [- \hat{i} + \frac{3 \hat{j}}{5}]$

$\vec{B} = \frac{μ_{0} i}{5 π} [- \hat{i} + \frac{7 \hat{j}}{24}]$

$\vec{B} = \frac{μ_{0} i}{7 π} [- 2 \hat{i} + \frac{3 \hat{j}}{5}]$

answer is B.

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Detailed Solution

From trigonometry

AP = 16/5 and PB - 915. Thus,

sin AOP - 4/5 and, cos AOP - 3/5

sin BOP = 3/5 and cos BOP - 4/5

$Magnetic field due to A is B_{A} = \frac{μ_{0} i}{8 π} which makes angle θ with n e g a t i v e x - a x i s a n d l i e s i n t h i r d q u a d r a n t . Magnetic field at O due to wire B is B_{B} = \frac{μ_{0} i}{6 π} which makes angle θ with y -axis and lies in second quadrant. I n c o m p o n e n t f o r m \vec{B_{A}} = \frac{μ_{0} i}{8 π} (- \frac{4}{5} i - \frac{3}{5} j) \vec{B_{B}} = \frac{μ_{0} i}{6 π} (- \frac{3}{5} i + \frac{4}{5} j) Thus, \vec{B_{A}} + \vec{B_{B}} = \frac{μ_{0} i}{5 π} (- i + \frac{7}{24} j)$