b
(b) Magnetic field at the centre due to one side
\({B_1} = \frac{{{\mu _0}}}{{4\pi }}.\frac{{2i\sin \theta }}{r}\)where \(r = a\cos \theta \)
So \({B_1} = \frac{{{\mu _0}}}{{4\pi }}.\frac{{2i\sin \theta }}{{a\cos \theta }} = \frac{{{\mu _0}i}}{{2\pi a}}\tan \theta \)
Hence net magnetic field
\({B_{net}} = n \times \frac{{{\mu _0}i}}{{2\pi a}}\tan \frac{\pi }{n}\).