A current $i$ is flowing in a straight conductor of length $L.$ The magnetic induction at a point on its axis at a distance $\frac {L}{4}$ from its centre will be
JEE MAIN 2013, Medium
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$B=\frac{\mu_{0} i}{4 \pi R}\left(\sin \theta_{1}+\sin \theta_{2}\right)$
$\sin \theta_{2}=\frac{L / 2}{\sqrt{(L / 2)^{2}+(L / 2)^{2}}}=\frac{1}{2(\sqrt{1 / 4+1 / 16})}=\frac{2}{\sqrt{5}}$
$\Rightarrow B=\frac{\mu_{0} i}{{4 \pi \frac{L}{4}}}\left(\frac{2 \times 4}{2 \sqrt{5}}\right) B=\frac{4 \mu_{0} i}{\sqrt{5} \pi L}$
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