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Physics-

General

Easy

Question

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

$\frac{μ_{0} i}{4 r}$
$\frac{μ_{0} i}{2 r}$
$\frac{μ_{0} i}{2 π r} (π + 1)$
$\frac{μ_{0} i}{2 π r} (π - 1)$

The correct answer is: $fraction numerator mu subscript 0 end subscript i over denominator 2 pi r end fraction open parentheses pi plus 1 close parentheses$

The magnitude of the magnetic field at point $O$ due to straight part of wire is
$B subscript 1 end subscript equals fraction numerator mu subscript 0 end subscript i over denominator 2 pi r end fraction$
$B subscript 1 end subscript$ is perpendicular to the plane of the page, directed upwards (right hand plam rule 1).
The field at the centre $O$ due to the current loop of radius $r$ is
$B subscript 2 end subscript equals fraction numerator mu subscript 0 end subscript i over denominator 2 r end fraction$

$B subscript 2 end subscript$ is also perpendicular to the page, directed upwards (right hand screw rule).
$B subscript 1 end subscript plus B subscript 2 end subscript equals fraction numerator mu subscript 0 end subscript i over denominator 2 r end fraction open parentheses fraction numerator 1 over denominator pi end fraction plus 1 close parentheses$
$equals fraction numerator mu subscript 0 end subscript i over denominator 2 pi r end fraction open parentheses pi plus 1 close parentheses$

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