MCQ
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
  • $\frac{{2{\mu _0}I}}{{\pi a}}$
  • B
    $\frac{{\sqrt 3 {\mu _0}I}}{{2\pi a}}$
  • C
    $\frac{{2\sqrt 2 {\mu _0}I}}{{\pi a}}$
  • D
    $\frac{{3\sqrt 3 {\mu _0}I}}{{2\pi a}}$

Answer

Correct option: A.
$\frac{{2{\mu _0}I}}{{\pi a}}$
a
$\mathrm{B}_{0}=\frac{2 \sqrt{2} \mu_{0} \mathrm{I}}{\pi \mathrm{b}}$

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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