A thin circular frame of radius $'a'$ is made of insulating material. A square loop is constructed with in it. If loop carrying current $I$ , then magnetic induction at geometrical centre $'O'$ will be
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$\mathrm{B}_{0}=\frac{2 \sqrt{2} \mu_{0} \mathrm{I}}{\pi \mathrm{b}}$
where, $\sin 45^{\circ}=\frac{\mathrm{b} / 2}{\mathrm{a}} \Rightarrow \mathrm{b}=\sqrt{2} \mathrm{a}$
Hence $\quad \mathrm{B}_{0}=\frac{2 \sqrt{2} \mu_{0} \mathrm{I}}{\pi(\sqrt{2} \mathrm{a})}=\frac{2 \mu_{0} \mathrm{I}}{\pi \mathrm{a}}$
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