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Question

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

Vedclass · Accepted Answer

$\frac{5{\mu _0}I	heta }{24\pi r}$

Vedclass · Answer

(A) The magnetic field due to a circular 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 $3r$. The current flows in such a way that the magnetic fields due to the arcs of radii $r$ and $3r$ are in one direction (say,into the page),while the field due to the arc of radius $2r$ is in the opposite direction (out of the page).Let $B_1$,$B_2$,and $B_3$ be the magnetic fields due to the arcs of radii $r$,$2r$,and $3r$ respectively.$B_1 = \frac{\mu_0 I 	heta}{4 \pi r}$ (into the page)$B_2 = \frac{\mu_0 I 	heta}{4 \pi (2r)}$ (out of the page)$B_3 = \frac{\mu_0 I 	heta}{4 \pi (3r)}$ (into the page)The net magnetic field $B$ at point $O$ 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]$$B = \frac{\mu_0 I 	heta}{4 \pi r} \left[ \frac{6 - 3 + 2}{6} ight]$$B = \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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