Figure shows a square loop ABCD with edge len | vedclass

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(b) 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{{{

Vedclass · Accepted Answer

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

Vedclass · Answer

(b) 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 ${i_1} + {i_2} = i$ $==>$ ${i_1} = \frac{{2i}}{3}$ 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 $i_1 = 2i_2. \,So \,(B_1 = B_2) > (B_3 = B_4)$

Hence net magnetic field at centre $O$
${B_{net}} = ({B_1} + {B_2}) - ({B_3} + {B_4})$
$ = 2 	imes \frac{{{\mu _0}}}{{4\pi }}.\frac{{2\sqrt 2 \, 	imes \left( {\frac{2}{3}i} ight)}}{a} - \frac{{{\mu _0}}}{{4\pi }}.\frac{{2\sqrt 2 \,\left( {\frac{i}{3}} ight) 	imes 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 $

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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