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Physics-

General

Easy

Question

Current $I$ is flowing in conductor shaped as shown in the figure. The radius of the curved part is $r$ and the length of straight portion is very large. The value of the magnetic field at the centre $O$ will be

$\frac{μ_{0} I}{4 π r} (\frac{3 π}{2} + 1)$
$\frac{μ_{0} I}{4 π r} (\frac{3 π}{2} - 1)$
$\frac{μ_{0} I}{4 π r} (\frac{π}{2} + 1)$
$\frac{μ_{0} I}{4 π r} (\frac{π}{2} - 1)$

The correct answer is: $fraction numerator mu subscript 0 end subscript I over denominator 4 pi r end fraction open parentheses fraction numerator 3 pi over denominator 2 end fraction plus 1 close parentheses$

$B subscript A end subscript equals 0$

$B subscript B end subscript equals blank fraction numerator mu subscript 0 end subscript over denominator 4 pi end fraction fraction numerator left parenthesis 2 pi minus pi divided by 2 right parenthesis I over denominator r end fraction blank circled times equals fraction numerator mu subscript 0 end subscript over denominator 4 pi end fraction fraction numerator 3 pi I over denominator 2 r end fraction$
$B subscript C end subscript equals fraction numerator mu subscript 0 end subscript I over denominator 4 pi r end fraction circled times$
So, net magnetic field at the centre
$equals B subscript A end subscript plus B subscript B end subscript plus blank B subscript C end subscript$
$equals 0 plus fraction numerator mu subscript 0 end subscript over denominator 4 pi end fraction fraction numerator 3 pi I over denominator 2 r end fraction plus fraction numerator mu subscript 0 end subscript I over denominator 4 pi r end fraction equals fraction numerator mu subscript 0 end subscript over denominator 4 pi end fraction fraction numerator I over denominator r end fraction open parentheses fraction numerator 3 pi over denominator 2 end fraction plus 1 close parentheses$

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