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(D) The magnetic field at the center of a circular arc of radius $R$ carrying current $I$ and subtending an angle $	heta$ at the center is given by $

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$3$

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(D) The magnetic field at the center of a circular arc of radius $R$ carrying current $I$ and subtending an angle $	heta$ at the center is given by $B = \frac{\mu_0 I 	heta}{4 \pi R}$.For a semicircular arc,the angle subtended at the center is $	heta = \pi$ radians.The straight segments of the conductor do not contribute to the magnetic field at the center $O$ because the point $O$ lies on the axis of these straight wires.Thus,the magnetic field at the center $O$ is due only to the semicircular arc:$B = \frac{\mu_0 I \pi}{4 \pi R} = \frac{\mu_0 I}{4 R}$.Given $I = 3 \, A$,$R = \frac{\pi}{10} \, m$,and $\mu_0 = 4 \pi 	imes 10^{-7} \, T \cdot m/A$:$B = \frac{4 \pi 	imes 10^{-7} 	imes 3}{4 	imes (\frac{\pi}{10})} = \frac{12 \pi 	imes 10^{-7}}{\frac{4 \pi}{10}} = \frac{12 \pi 	imes 10^{-7} 	imes 10}{4 \pi} = 3 	imes 10^{-6} \, T$.Since $1 \, \mu T = 10^{-6} \, T$,we have $B = 3 \, \mu T$.

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