A loop ABCDA,carrying current I=12 A,is plac | vedclass

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(D) The magnetic field due to the straight segments $AB$ and $CD$ at the center $O$ is zero because the line of action of these segments passes throug

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$1$

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(D) The magnetic field due to the straight segments $AB$ and $CD$ at the center $O$ is zero because the line of action of these segments passes through the center $O$.The magnetic field due to a semi-circular arc of radius $R$ carrying current $I$ at its center is given by $B = \frac{\mu_0 I}{4R}$.The magnetic field due to the outer semi-circular arc of radius $R_1$ is $B_1 = \frac{\mu_0 I}{4R_1}$ (directed into the plane).The magnetic field due to the inner semi-circular arc of radius $R_2$ is $B_2 = \frac{\mu_0 I}{4R_2}$ (directed out of the plane).The resultant magnetic field at $O$ is $B_0 = |B_2 - B_1| = \frac{\mu_0 I}{4} \left( \frac{1}{R_2} - \frac{1}{R_1} ight)$.Substituting the given values: $I = 12 \ A$,$R_1 = 6 \pi \ m$,$R_2 = 4 \pi \ m$,and $\mu_0 = 4 \pi 	imes 10^{-7} \ Tm \ A^{-1}$.$B_0 = \frac{4 \pi 	imes 10^{-7} 	imes 12}{4} \left( \frac{1}{4 \pi} - \frac{1}{6 \pi} ight)$$B_0 = 12 \pi 	imes 10^{-7} \left( \frac{3 - 2}{12 \pi} ight)$$B_0 = 12 \pi 	imes 10^{-7} \left( \frac{1}{12 \pi} ight) = 1 	imes 10^{-7} \ T$.Comparing this with $k 	imes 10^{-7} \ T$,we get $k = 1$.

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