The wire loop carries current i as shown in the figure. The magnetic field at the centre O is :

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zero

$\frac{μ_{0} i}{2} (\frac{1}{R_{1}} - \frac{1}{R_{2}})$

$\frac{μ_{0} i}{4} (\frac{1}{R_{1}} - \frac{1}{R_{2}})$

$\frac{μ_{0} i}{4} (\frac{1}{R_{1}} + \frac{1}{R_{2}})$

answer is B.

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Detailed Solution

${\bar{B}}_{0} = {\bar{B}}_{1} + {\bar{B}}_{2}$

$\begin{array}{l} = \frac{{μ_{0} i}^{}}{4 R_{1}} - \frac{μ_{0}^{} i}{4 R_{2}} \\ {\bar{B}}_{0} = \frac{μ_{0} i}{4} [\frac{1}{R_{1}} - \frac{1}{R_{2}}] ⊙ \end{array}$

B at O due to PQ and RS is zero

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