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 magnetic field at the centre of the loop assuming uniform wire is

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$\frac{{\sqrt 2 \,{\mu _0}i}}{{3\pi \,a}} \odot$

$\frac{{\sqrt 2 \,{\mu _0}i}}{{3\pi \,a}} \otimes$

$\frac{{\sqrt 2 \,{\mu _0}i}}{{\pi \,a}} \odot$

$\frac{{\sqrt 2 \,{\mu _0}i}}{{\pi \,a}} \otimes$

answer is B.

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Detailed Solution

According to question resistance of wire ADC is twice that of wire ABC. Hence current flows through ADC is half that of ABC i.e. $\frac{{{i_2}}}{{{i_1}}} = \frac{1}{2}$ .
Also ${i_1} + {i_2} = i$ $\Rightarrow {i_1} = \frac{{2i}}{3}\,and\,{i_2} = \frac{i}{3}$
Magnetic field at centre O due to wire AB and BC (part 1 and 2) ${B_1} = {B_2} = \frac{{{\mu _0}}}{{4\pi }}.\frac{{2{i_1}\sin {{45}^o}}}{{a/2}} \otimes = \frac{{{\mu _0}}}{{4\pi }}.\frac{{2\sqrt 2 \,{i_1}}}{a} \otimes$
and magnetic field at centre O due to wires AD and DC (i.e. part 3 and 4) ${B_3} = {B_4} = \frac{{{\mu _0}}}{{4\pi }}\frac{{2\sqrt 2 \,{i_2}}}{a} \odot$
Also, i₁ = 2i₂. So (B₁ = B₂) > (B₃ = B₄)
Hence net magnetic field at centre O
${B_{net}} = ({B_1} + {B_2}) - ({B_3} + {B_4})$
$= 2 \times \frac{{{\mu _0}}}{{4\pi }}.\frac{{2\sqrt 2 \, \times \left( {\frac{2}{3}i} \right)}}{a} - \frac{{{\mu _0}}}{{4\pi }}.\frac{{2\sqrt 2 \,\left( {\frac{i}{3}} \right) \times 2}}{a}$
$= \frac{{{\mu _0}}}{{4\pi }}.\frac{{4\sqrt 2 \,i}}{{3a}}(2 - 1)\, \otimes = \frac{{\sqrt 2 \,{\mu _0}i}}{{3\pi \,a}} \otimes$