Question
Find magnetic field at $O$
✓
Answer
(a) ${B_1} = {B_3} = {B_5} = 0$
${B_2} = \frac{{{\mu _0}}}{{4\pi }}.\frac{{\theta i}}{{3r}} \otimes ,\;{B_4} = \frac{{{\mu _0}}}{{4\pi }}.\frac{{\theta i}}{{2r}}$ $\odot$
and ${B_6} = \frac{{{\mu _0}}}{{4\pi }}.\frac{{\theta i}}{r} \otimes $
$\therefore $ Net magnetic field at $O$,
${B_{net}} = {B_2} - {B_4} + {B_6}$ $ = \frac{{{\mu _0}}}{{4\pi }}.\frac{{\theta i}}{r}\left( {\frac{1}{3} - \frac{1}{2} + 1} \right) = \frac{{5{\mu _0}\theta i}}{{24\pi r}}$
${B_2} = \frac{{{\mu _0}}}{{4\pi }}.\frac{{\theta i}}{{3r}} \otimes ,\;{B_4} = \frac{{{\mu _0}}}{{4\pi }}.\frac{{\theta i}}{{2r}}$ $\odot$
and ${B_6} = \frac{{{\mu _0}}}{{4\pi }}.\frac{{\theta i}}{r} \otimes $
$\therefore $ Net magnetic field at $O$,
${B_{net}} = {B_2} - {B_4} + {B_6}$ $ = \frac{{{\mu _0}}}{{4\pi }}.\frac{{\theta i}}{r}\left( {\frac{1}{3} - \frac{1}{2} + 1} \right) = \frac{{5{\mu _0}\theta i}}{{24\pi r}}$
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