A long straight conductor, carrying a current I, is bent into the shape shown in the figure. The radius of the circular loop is r. The magnetic field at the centre of the loop is

\frac{μ_{0} i}{2 r} [1 + \frac{1}{π}]

into the page

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\frac{μ_{0} i}{2 r} [1 - \frac{1}{π}]

out of the page

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\frac{μ_{0} i}{2 r} [1 - \frac{1}{π}]

into the page

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\frac{μ_{0} i}{2 r} [1 + \frac{1}{π}]

out of the page

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Solution

The correct option is C $\frac{μ_{0} i}{2 r} [1 - \frac{1}{π}]$ into the page

$B_{1} = \frac{μ_{0} i}{2 r}$ for Circular loop $θ$
$B_{2} = \frac{μ_{0} i}{2 π r}$ for straight wire
$B = B_{1} - B_{2} = \frac{μ_{0} i}{2 r} [1 - \frac{1}{π}]$ in to the page

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A long straight conductor, carrying a current I, is bent into the shape shown in the figure. The radius of the circular loop is r. The magnetic field at the centre of the loop is

The correct option is C μ0i2r[1−1π] into the page B1=μ0i2r for Circular loop θ B2=μ0i2πr for straight wire B=B1−B2=μ0i2r[1−1π] in to the page

The correct option is C $\frac{μ_{0} i}{2 r} [1 - \frac{1}{π}]$ into the page

$B_{1} = \frac{μ_{0} i}{2 r}$ for Circular loop $θ$
$B_{2} = \frac{μ_{0} i}{2 π r}$ for straight wire
$B = B_{1} - B_{2} = \frac{μ_{0} i}{2 r} [1 - \frac{1}{π}]$ in to the page