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(C) The magnetic field at the center $O$ due to the circular loop of radius $r$ is given by $B_{	ext{loop}} = \frac{{\mu _0 i}}{{2r}}$,directed perpe

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$\frac{{\mu _0 i}}{{2\pi r}}(\pi + 1)$

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(C) The magnetic field at the center $O$ due to the circular loop of radius $r$ is given by $B_{	ext{loop}} = \frac{{\mu _0 i}}{{2r}}$,directed perpendicularly outwards (using the right-hand thumb rule).The magnetic field at the center $O$ due to the long straight wire at a distance $r$ is given by $B_{	ext{wire}} = \frac{{\mu _0 i}}{{2\pi r}}$,also directed perpendicularly outwards.Since both magnetic fields are in the same direction,the net magnetic field $B_{	ext{net}}$ is the sum of the two:$B_{	ext{net}} = B_{	ext{loop}} + B_{	ext{wire}}$$B_{	ext{net}} = \frac{{\mu _0 i}}{{2r}} + \frac{{\mu _0 i}}{{2\pi r}}$$B_{	ext{net}} = \frac{{\mu _0 i}}{{2\pi r}}(\pi + 1)$

$A$ part of a long wire carrying a current $i$ is bent into a circle of radius $r$ as shown in the figure. The net magnetic field at the centre $O$ of the circular loop is

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