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Question

(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 	het

Vedclass · Accepted Answer

$\frac{\mu_{0} I}{12}\left[\frac{1}{R_{1}}-\frac{1}{R_{2}}\right]$

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 $B = \frac{\mu_{0} I 	heta}{4 \pi R}$.Here,the angle subtended is $	heta = 60^{\circ} = \frac{\pi}{3} 	ext{ radians}$.The magnetic field due to the inner arc of radius $R_1$ is $B_1 = \frac{\mu_{0} I (\pi/3)}{4 \pi R_1} = \frac{\mu_{0} I}{12 R_1}$ (directed into the page).The magnetic field due to the outer arc of radius $R_2$ is $B_2 = \frac{\mu_{0} I (\pi/3)}{4 \pi R_2} = \frac{\mu_{0} I}{12 R_2}$ (directed out of the page).The straight segments do not contribute to the magnetic field at $O$ because the position vector of $O$ is collinear with the current elements.The net magnetic field at $O$ is $B = B_1 - B_2 = \frac{\mu_{0} I}{12} \left( \frac{1}{R_1} - \frac{1}{R_2} ight)$.

$A$ conducting wire carrying current $I$ is arranged as shown in the figure. Find the magnetic field at point $O$.

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