Figure shows three cases: in all cases the circular part has radius $r$ and straight ones are infinitely long. For the same current the ratio of field $B$ at center $P$ in the three cases $B_{1} : B_{2} : B_{3}$ is :

(- \frac{π}{2}) : (\frac{π}{2}) : (\frac{3 π}{4} - \frac{1}{2})

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(- \frac{π}{2} + 1) : (\frac{π}{2} + 1) : (\frac{3 π}{4} + \frac{1}{2})

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(- \frac{π}{2}) : (\frac{π}{2}) : (\frac{3 π}{4})

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(- \frac{π}{2} - 1) : (\frac{π}{2} - \frac{1}{4}) : (\frac{3 π}{4} + \frac{1}{2})

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Solution

The correct option is A $(- \frac{π}{2}) : (\frac{π}{2}) : (\frac{3 π}{4} - \frac{1}{2})$
$_{1} = \frac{μ_{0} I}{4 r} ⨀$
$_{2} = \frac{μ_{0} I}{4 r} ⨂$
$_{3} = \frac{μ_{0} I}{2 r} (\frac{3}{4}) ⨂ + \frac{μ_{0} I}{4 π r} ⨀$
$= \frac{μ_{0} I}{4 r} [\frac{3}{2} ⨂ + \frac{1}{π} ⨀]$
$= \frac{μ_{0} I}{4 r} [\frac{3 π - 2}{2 π}] ⨂ = \frac{μ_{0} I}{4 r} (\frac{2}{π}) (\frac{3 π}{4} - \frac{1}{2}) ⨂$
$B_{1} : B_{2} : B_{3} = - (\frac{π}{2}) : (\frac{π}{2}) : (\frac{3 π}{4} - \frac{1}{2})$

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