A current-carrying loop ABCD has two circular | vedclass

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(C) The magnetic field at the centre $O$ due to a circular arc of radius $R$ subtending an angle $	heta$ (in radians) is given by $B = \frac{\mu_0 i

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$1$

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(C) The magnetic field at the centre $O$ due to a circular arc of radius $R$ subtending an angle $	heta$ (in radians) is given by $B = \frac{\mu_0 i 	heta}{4 \pi R}$.Here,$	heta = 30^{\circ} = \frac{\pi}{6} 	ext{ radians}$.The straight segments $AB$ and $CD$ pass through the centre $O$,so the magnetic field due to them is zero $(B_{AB} = 0, B_{CD} = 0)$.The magnetic field due to arc $AD$ (radius $R_1 = 0.01 	ext{ m}$) is $B_{AD} = \frac{\mu_0 i}{4 \pi R_1} 	imes \frac{\pi}{6} = \frac{\mu_0 i}{24 R_1}$ (directed outwards,$\odot$).The magnetic field due to arc $BC$ (radius $R_2 = 0.02 	ext{ m}$) is $B_{BC} = \frac{\mu_0 i}{4 \pi R_2} 	imes \frac{\pi}{6} = \frac{\mu_0 i}{24 R_2}$ (directed inwards,$\otimes$).The net magnetic field is $B_{net} = B_{AD} - B_{BC} = \frac{\mu_0 i}{24} \left( \frac{1}{R_1} - \frac{1}{R_2} ight)$.Substituting the values: $B_{net} = \frac{4 \pi 	imes 10^{-7} 	imes (1.2 / \pi)}{24} \left( \frac{1}{0.01} - \frac{1}{0.02} ight) = \frac{4.8 	imes 10^{-7}}{24} (100 - 50) = 0.2 	imes 10^{-7} 	imes 50 = 10 	imes 10^{-7} 	ext{ T} = 1 \mu 	ext{T}$.

Column $A$	Column $B$
$A) \frac{B}{4}$	$i) (0, 0)$
$B) \frac{B}{2}$	$ii) (a, 0)$
$C) \frac{2B}{3}$	$iii) (2a, 0)$
$D) 2B$	$iv) (3a, 0)$

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