Magnetic field due to currentMedium

Question

Two infinitely long straight parallel wires perpendicular to the plane of the paper are 5 m apart. One of the wires, P carries current I out of the plane of the paper and the other, Q carries the current I into the plane of paper. The magnetic field B at the origin O of the coordinate system with x and y axes as perpendicular and parallel to PQ, respectively, is [Given OP = 4 m and OQ = 3 m]

Options

A.(μ0I/2π)(î − (3/5)ĵ)
B.(μ0I/5π)(î − (7/24)ĵ)
C.(μ0I/5π)(−î + (3/8)ĵ)
D.(μ0I/24π)(2î + 3ĵ)

Solution

[Key Concept: The field of a long straight wire is μ0I/(2πr), perpendicular to the radius; fields of several wires add as vectors.]

$$PM = \frac{16}{5}\text{ }m\& QM = \frac{9}{5}\text{ }m$$

$$\begin{matrix} & {\overrightarrow{B}}_{P} = \frac{\mu_{0}I}{8\pi}\left\lbrack \frac{4}{5}\widehat{i} + \frac{3}{5}\widehat{j} \right\rbrack \\ & {\overrightarrow{B}}_{Q} = \frac{\mu_{0}I}{6\pi}\left\lbrack \frac{3}{5}\widehat{i} - \frac{4}{5}\widehat{j} \right\rbrack \\ & \ \therefore{\overrightarrow{B}}_{\text{net~}} = \frac{\mu_{0}I}{\pi}\left\lbrack \frac{1}{10}\widehat{i} + \frac{3}{40}\widehat{j} + \frac{1}{10}\widehat{i} - \frac{2}{15}\widehat{j} \right\rbrack \\ & \ = \frac{\mu_{0}I}{\pi}\left\lbrack \frac{1}{5}\widehat{i} - \frac{7}{120}\widehat{j} \right\rbrack = \frac{\mu_{0}I}{5\pi}\left\lbrack \widehat{i} - \frac{7}{24}\widehat{j} \right\rbrack \end{matrix}$$

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