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

Vedclass · Accepted Answer

$\frac{5\mu_0 I	heta}{24\pi r}$

Vedclass · Answer

(A) The magnetic field at the center of 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 $3r$.$1$. For the first arc (radius $r$),the current flows in a direction such that the magnetic field at $O$ is directed into the page (using the right-hand rule). Let this be $B_1 = \frac{\mu_0 I 	heta}{4\pi r}$.$2$. For the second arc (radius $2r$),the current flows in the opposite direction,so the magnetic field is directed out of the page. Let this be $B_2 = \frac{\mu_0 I 	heta}{4\pi (2r)} = \frac{\mu_0 I 	heta}{8\pi r}$.$3$. For the third arc (radius $3r$),the current flows in the same direction as the first arc,so the magnetic field is directed into the page. Let this be $B_3 = \frac{\mu_0 I 	heta}{4\pi (3r)} = \frac{\mu_0 I 	heta}{12\pi r}$.The net magnetic field $B_{net}$ at $O$ is $B_1 - B_2 + B_3$:$B_{net} = \frac{\mu_0 I 	heta}{4\pi} \left( \frac{1}{r} - \frac{1}{2r} + \frac{1}{3r} ight)$$B_{net} = \frac{\mu_0 I 	heta}{4\pi r} \left( 1 - \frac{1}{2} + \frac{1}{3} ight)$$B_{net} = \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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