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(C) The total current $I$ divides into two parallel paths $	ext{ABC}$ and $	ext{ADC}$.Since the resistance of $	ext{ABC}$ is $r$ and $	ext{ADC}$ i

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$\frac{\sqrt{2} \mu_0 I}{3 \pi a}$

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(C) The total current $I$ divides into two parallel paths $	ext{ABC}$ and $	ext{ADC}$.Since the resistance of $	ext{ABC}$ is $r$ and $	ext{ADC}$ is $2r$,the currents $i_1$ and $i_2$ are inversely proportional to the resistances:$i_1 = I \cdot \frac{2r}{r + 2r} = \frac{2I}{3}$ (through $	ext{ABC}$)$i_2 = I \cdot \frac{r}{r + 2r} = \frac{I}{3}$ (through $	ext{ADC}$)The magnetic field at the centre due to a finite wire of length $a$ at distance $d = a/2$ is $B = \frac{\mu_0 i}{4 \pi d} (\sin 	heta_1 + \sin 	heta_2)$. For a square,$	heta_1 = 	heta_2 = 45^\circ$,so $B = \frac{\mu_0 i}{4 \pi (a/2)} (\frac{1}{\sqrt{2}} + \frac{1}{\sqrt{2}}) = \frac{\mu_0 i}{\sqrt{2} \pi a}$.The field due to $	ext{ABC}$ (carrying $2I/3$) is $B_1 = 2 	imes \frac{\mu_0 (2I/3)}{\sqrt{2} \pi a} = \frac{2\sqrt{2} \mu_0 I}{3 \pi a}$ (into the page).The field due to $	ext{ADC}$ (carrying $I/3$) is $B_2 = 2 	imes \frac{\mu_0 (I/3)}{\sqrt{2} \pi a} = \frac{\sqrt{2} \mu_0 I}{3 \pi a}$ (out of the page).The resultant field is $B_{net} = B_1 - B_2 = \frac{2\sqrt{2} \mu_0 I}{3 \pi a} - \frac{\sqrt{2} \mu_0 I}{3 \pi a} = \frac{\sqrt{2} \mu_0 I}{3 \pi a}$.

The figure shows a current-carrying square loop $\text{ABCD}$ of edge length '$a$' lying in a plane. If the resistance of the $\text{ABC}$ part is $r$ and that of the $\text{ADC}$ part is $2r$,then the magnitude of the resultant magnetic field at the centre of the square loop is:

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