A straight conductor carrying current i splits into two parts as shown in the figure. The radius of the circular loop is R. The total magnetic field at the centre P at the loop is :

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$3 μ_{0} i / 32 R$ , outword

$3 μ_{0} i / 32 R$ , inward

Zero

$\frac{μ_{0} i}{2 R}$ , inward

answer is A.

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Detailed Solution

$\frac{i_{1}}{i_{2}} = \frac{\frac{1}{4} (2 π R)}{\frac{3}{4} (2 π R)}$

$\Rightarrow i_{2} = 3 i_{1}$

Net magnetic field, $= \frac{μ_{0} i_{1}}{2 R} (\frac{θ_{1}}{2 π}) - \frac{μ_{0} i_{2}}{2 R} (\frac{θ_{2}}{2 π})$

$= \frac{μ_{0} i_{1}}{2 R} (\frac{3 π}{2 \times 2 π}) - \frac{μ_{0} i_{2}}{2 R} (\frac{π}{2 \times 2 π})$

$= \frac{μ_{0}}{2 R} [\frac{3 i_{1}}{4} - \frac{i_{2}}{4}]$

$= \frac{μ_{0}}{2 R} [\frac{3 i_{1}}{4} - \frac{3 i_{1}}{4}] = 0$

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