A long straight conductor, carrying a current I, is bent to form an almost complete circular loop of radius $r$ on it. The magnetic field at the center of the loop:

is directed into the paper

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is directed out of the paper

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has magnitude

\frac{μ_{0} I}{2 r} (1 - \frac{1}{π})

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has magnitude

\frac{μ_{0} I}{2 r} (1 + \frac{1}{π})

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Solution

The correct option is D has magnitude $\frac{μ_{0} I}{2 r} (1 - \frac{1}{π})$
The magnetic field at the center of coil due to current carrying coil is,
$B_{1} = \frac{μ_{0} I}{2 r}$ upwards.
Similarly the magnetic field at the same point due to current carrying straight wire at a distance $r$ is ,
$B_{2} = \frac{μ_{0} I}{2 π r}$ downwards.
Hence the net field upwards = $\frac{μ_{0} I}{2 r} (1 - \frac{1}{π})$

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A long straight conductor, carrying a current I, is bent to form an almost complete circular loop of radius r on it. The magnetic field at the center of the loop:

A long straight conductor, carrying a current I, is bent to form an almost complete circular loop of radius $r$ on it. The magnetic field at the center of the loop: