In the given figure net magnetic field at O will be - Vidyadip

MCQ

In the given figure net magnetic field at $O$ will be

A
$\frac{{2{\mu _0}i}}{{3\pi a}}\sqrt {4 - {\pi ^2}} $
✓
$\frac{{{\mu _0}i}}{{3\pi a}}\sqrt {4 + {\pi ^2}} $
C
$\frac{{2{\mu _0}i}}{{3\pi {a^2}}}\sqrt {4 + {\pi ^2}} $
D
$\frac{{2{\mu _0}i}}{{3\pi a}}\sqrt {(4 - {\pi ^2})} $

✓

Answer

Correct option: B.

$\frac{{{\mu _0}i}}{{3\pi a}}\sqrt {4 + {\pi ^2}} $

b
(b) Magnetic field at $O$ due to Part $(1)$ : ${B_1} = 0$

Part $(2)$: ${B_2} = \frac{{{\mu _0}}}{{4\pi }}.\frac{{\pi \,i}}{{(a/2)}} \otimes $(along $-Z-$axis)

Part $(3)$: ${B_3} = \frac{{{\mu _0}}}{{4\pi }}.\frac{i}{{(a/2)}}\left( \downarrow \right)$(along $-Y-$axis)

Part $(4)$: ${B_4} = \frac{{{\mu _0}}}{{4\pi }}.\frac{{\pi i}}{{(3a/2)}}\odot$(along $+Z-$axis)

Part $(5)$: ${B_5} = \frac{{{\mu _0}}}{{4\pi }}.\frac{i}{{(3a/2)}}\left( \downarrow \right)$(along -$Y$ $-$ axis) ${B_2} - {B_4} = \frac{{{\mu _0}}}{{4\pi }}.\frac{{\pi i}}{a}\left( {2 - \frac{2}{3}} \right) = \frac{{{\mu _0}i}}{{3a}} \otimes $(along -$Z$ $-$ axis) ${B_3} + {B_5} = \frac{{{\mu _0}}}{{4\pi }}.\frac{1}{a}\left( {2 + \frac{2}{3}} \right) = \frac{{8{\mu _0}i}}{{12\pi a}}\left( \downarrow \right)$ (along -$Y$$ -$ axis)

Hence net magnetic field ${B_{net}} = \sqrt {{{({B_2} - {B_4})}^2} + {{({B_3} + {B_5})}^2}} $$ = \frac{{{\mu _0}i}}{{3\pi a}}\sqrt {{\pi ^2} + 4} $

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