current r is flowing in a straight conductor of length l. The magnetic induction at a point distant l /4 from its centre will be(fig)

See fig. We know that magnetic induction B at a distance r from a long straight conductor carrying a current i is given by
 $\mathrm{B}=\frac{{\mathrm{\mu }}_0\mathrm{i}}{4\mathrm{\pi r}}[\sin \mathrm{\alpha }+\sin \mathrm{\beta }]$
Here $\mathrm{\alpha }=\mathrm{\alpha },\mathrm{\beta }=\mathrm{\alpha }\text{ and }\mathrm{r}=\mathrm{l}/4$
$\begin{array}{r}\therefore (\sin \mathrm{\alpha }+\sin \mathrm{\beta })=2\sin \mathrm{\alpha }=2\frac{\mathrm{l}/2}{\sqrt{5}\mathrm{l}/4}=\frac{4}{\sqrt{5}}\\ \therefore \mathrm{B}=\frac{{\mathrm{\mu }}_0\mathrm{i}}{4\mathrm{\pi }(\mathrm{l}/4)}\times \frac{4}{\sqrt{5}}=\frac{4{\mathrm{\mu }}_0\mathrm{i}}{\sqrt{5}\mathrm{\pi l}}\end{array}$