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