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 B₁: B₂: B₃is

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

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

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

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

answer is A.

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Detailed Solution

${\vec{B}}_{1} = \frac{μ_{0} I}{4 r} ⊙; {\vec{B}}_{2} = \frac{μ_{0} I}{4 r} \otimes {\vec{B}}_{3} = \frac{μ_{0} I}{2 r} (\frac{3}{4}) \otimes + \frac{μ_{0} I}{4 π r} ⊙ = \frac{μ_{0} I}{4 r} [\frac{3}{2} \otimes + \frac{1}{π} ⊙] = \frac{μ_{0} I}{4 r} [\frac{3 π - 2}{2 π}] \otimes = \frac{μ_{0} I}{4 r} (\frac{2}{π}) (\frac{3 π}{4} - \frac{1}{2}) \otimes B_{1} : B_{2} : B_{3} = - (\frac{π}{2}) : (\frac{π}{2}) : (\frac{3 π}{4} - \frac{1}{2})$