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(D) The straight wire segments $ab$ and $cd$ do not produce any magnetic field at point $O$ because the point $O$ lies on the axis of these wires.For

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$1 \, \mu T$

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(D) The straight wire segments $ab$ and $cd$ do not produce any magnetic field at point $O$ because the point $O$ lies on the axis of these wires.For the arc $bc$ (radius $R_1 = 2 \, cm = 2 	imes 10^{-2} \, m$), the magnetic field $B_1$ at $O$ is directed inward $(\otimes)$ and is given by:$B_1 = \frac{\mu_0 I}{4 \pi R_1} 	heta = \frac{10^{-7} 	imes I 	imes 	heta}{R_1}$$B_1 = \frac{10^{-7} 	imes (1.2 / \pi) 	imes (30^\circ 	imes \pi / 180^\circ)}{2 	imes 10^{-2}} = \frac{10^{-7} 	imes 1.2 	imes (1/6)}{2 	imes 10^{-2}} = 10^{-6} \, T = 1000 \, nT$ (inward).For the arc $ad$ (radius $R_2 = 1 \, cm = 1 	imes 10^{-2} \, m$), the magnetic field $B_2$ at $O$ is directed outward $(\odot)$ and is given by:$B_2 = \frac{\mu_0 I}{4 \pi R_2} 	heta = \frac{10^{-7} 	imes (1.2 / \pi) 	imes (30^\circ 	imes \pi / 180^\circ)}{1 	imes 10^{-2}} = \frac{10^{-7} 	imes 1.2 	imes (1/6)}{1 	imes 10^{-2}} = 2 	imes 10^{-6} \, T = 2000 \, nT$ (outward).The net magnetic field at $O$ is $B_{net} = B_2 - B_1 = 2000 \, nT - 1000 \, nT = 1000 \, nT = 1 \, \mu T$.

The magnitude of the magnetic field at $O$ due to a current-carrying loop as shown in the figure is, where $O$ is the center of two circular portions with radii $1 \, cm$ and $2 \, cm$ respectively. (Take the value of current $I = \frac{1.2}{\pi} \, A$)

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