Two concentric coplanar circular loops of rad | vedclass

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Question

(d) Magnetic field at centre due to smaller loop
${B_1} = \frac{{{\mu _0}}}{{4\pi }}.\frac{{2\pi {i_1}}}{{{r_1}}}$..... $(i)$
Due to Bigger loop ${B

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(d) Magnetic field at centre due to smaller loop
${B_1} = \frac{{{\mu _0}}}{{4\pi }}.\frac{{2\pi {i_1}}}{{{r_1}}}$..... $(i)$
Due to Bigger loop ${B_2} = \frac{{{\mu _0}}}{{4\pi }}.\frac{{2\pi {i_2}}}{{{r_2}}}$ So net magnetic field at centre
$B = {B_1} - {B_2} = \frac{{{\mu _0}}}{{4\pi }} 	imes 2\pi \left( {\frac{{{i_1}}}{{{r_1}}} - \frac{{{i_2}}}{{{r_2}}}} ight)$
According to question $B = \frac{1}{2} 	imes {B_1}$
$ \Rightarrow \frac{{{\mu _0}}}{{4\pi }}.2\pi \left( {\frac{{{i_1}}}{{{r_1}}} - \frac{{{i_2}}}{{{r_2}}}} ight) = \frac{1}{2} 	imes \frac{{{\mu _0}}}{{4\pi }}.\frac{{2\pi {i_1}}}{{{r_1}}}$ $\frac{{{i_1}}}{{{r_1}}} - \frac{{{i_2}}}{{{r_2}}} = \frac{{{i_1}}}{{2{r_1}}} \Rightarrow \frac{{{i_1}}}{{2{r_1}}} = \frac{{{i_2}}}{{{r_2}}} \Rightarrow \frac{{{i_1}}}{{{i_2}}} = 1$ $\{ {r_2} = 2{r_1}\} $

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