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(B) According to the question,the resistance of wire $ADC$ is twice that of wire $ABC$. Hence,the current flowing through $ADC$ is half that of $ABC$,

Vedclass · Accepted Answer

$\frac{\sqrt{2} \mu_0 i}{3\pi a} \otimes$

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(B) According to the question,the resistance of wire $ADC$ is twice that of wire $ABC$. Hence,the current flowing through $ADC$ is half that of $ABC$,i.e.,$i_2 = i_1 / 2$. Since $i_1 + i_2 = i$,we have $i_1 + i_1/2 = i$,which gives $i_1 = 2i/3$ and $i_2 = i/3$. The magnetic field at the centre $O$ due to wires $AB$ and $BC$ (carrying current $i_1$) is $B_{ABC} = 2 	imes \frac{\mu_0}{4\pi} \frac{2i_1 \sin 45^\circ}{a/2} = \frac{\mu_0}{4\pi} \frac{4\sqrt{2} i_1}{a} \otimes$. Substituting $i_1 = 2i/3$,we get $B_{ABC} = \frac{\mu_0}{4\pi} \frac{8\sqrt{2} i}{3a} \otimes$. The magnetic field at the centre $O$ due to wires $AD$ and $DC$ (carrying current $i_2$) is $B_{ADC} = 2 	imes \frac{\mu_0}{4\pi} \frac{2i_2 \sin 45^\circ}{a/2} = \frac{\mu_0}{4\pi} \frac{4\sqrt{2} i_2}{a} \odot$. Substituting $i_2 = i/3$,we get $B_{ADC} = \frac{\mu_0}{4\pi} \frac{4\sqrt{2} i}{3a} \odot$. The net magnetic field at the centre $O$ is $B_{net} = B_{ABC} - B_{ADC} = \frac{\mu_0}{4\pi} \frac{4\sqrt{2} i}{3a} (2 - 1) \otimes = \frac{\sqrt{2} \mu_0 i}{3\pi a} \otimes$.

The figure shows a square loop $ABCD$ with edge length $a$. The resistance of the wire $ABC$ is $r$ and that of $ADC$ is $2r$. The value of the magnetic field at the centre of the loop is:

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