The magnetic field B at the centre of a circular coil of radius r is π times that due to a long straight wire at a distance r from it, for equal currents. The diagram shows three cases: in all cases the circular part has radius r and straight ones are infinitely long. For the same current, the field B at centre P in cases 1,2,3 has the ratio :

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

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

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

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

answer is A.

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Detailed Solution

$\begin{array}{l} (1) \begin{array}{c} B_{1} = \frac{μ_{0} I}{4 πr} = B_{3} \\ \otimes \end{array} \\ ⊙ B_{2} = \frac{μ_{0} I}{4 r} \\ {\vec{B}}_{net} = {\vec{B}}_{1} + {\vec{B}}_{2} + {\vec{B}}_{3} \\ B_{net} = \frac{μ_{0} I}{4 r} ⊙ \dots (i) \end{array}$

$\begin{array}{l} B_{1} = B_{3} = 0 {\vec{B}}_{net} = {\vec{B}}_{1} + {\vec{B}}_{2} \\ + {\vec{B}}_{3} \\ \otimes B_{2} = \frac{μ_{0} I}{4 R} \\ = \frac{μ_{0} I}{4 r} \otimes \dots (ii) \end{array}$

$(3) B_{1} = 0$

$\otimes B_{2} = \frac{μ_{0} I}{4 πr} (\frac{3 π}{2}) = (\frac{3 μ_{0} I}{8 r})$

$⊙ B_{3} = \frac{μ_{0} I}{4 π r}$

$\begin{array}{l} {\vec{B}}_{net} = \frac{μ_{0} I}{4 r} (\frac{3}{2} - \frac{1}{π}) \otimes \dots (iii) \\ B_{(1)} : B_{(2)} : B_{(3)} \\ - 1 : 1 : (\frac{3}{2} - \frac{1}{π}) \\ \Rightarrow \frac{- π}{2} : \frac{π}{2} : (\frac{3 π}{4} - \frac{1}{2}) \end{array}$