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(A) The magnetic field due to an arc of radius $R$ subtending an angle $	heta$ at the center is given by $B = \frac{\mu_0 I 	heta}{4 \pi R}$.In the

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$\frac {5\mu _0I	heta }{24\,\pi r}$

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(A) The magnetic field due to an arc of radius $R$ subtending an angle $	heta$ at the center is given by $B = \frac{\mu_0 I 	heta}{4 \pi R}$.In the given figure,there are three arcs with radii $r$,$2r$,and $3r$ respectively,all subtending the same angle $	heta$ at point $O$.Using the right-hand rule,the magnetic field due to the arc of radius $r$ is directed into the page,the arc of radius $2r$ is directed out of the page,and the arc of radius $3r$ is directed into the page.Let $B_1, B_2, B_3$ be the magnetic fields due to the arcs of radii $r, 2r, 3r$ respectively.$B_1 = \frac{\mu_0 I 	heta}{4 \pi r}$ (inward)$B_2 = \frac{\mu_0 I 	heta}{4 \pi (2r)} = \frac{\mu_0 I 	heta}{8 \pi r}$ (outward)$B_3 = \frac{\mu_0 I 	heta}{4 \pi (3r)} = \frac{\mu_0 I 	heta}{12 \pi r}$ (inward)The net magnetic field $B$ at $O$ is $B = B_1 - B_2 + B_3$.$B = \frac{\mu_0 I 	heta}{4 \pi} \left[ \frac{1}{r} - \frac{1}{2r} + \frac{1}{3r} ight]$$B = \frac{\mu_0 I 	heta}{4 \pi r} \left[ 1 - \frac{1}{2} + \frac{1}{3} ight]$$B = \frac{\mu_0 I 	heta}{4 \pi r} \left[ \frac{6 - 3 + 2}{6} ight] = \frac{\mu_0 I 	heta}{4 \pi r} \left[ \frac{5}{6} ight] = \frac{5 \mu_0 I 	heta}{24 \pi r}$.

Shown in the figure is a conductor carrying a current $I$. The magnetic field intensity at the point $O$ (common centre of all the three arcs) is

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