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A current carrying loop ABCDA is carrying a current I. O is the centre of the semi circle ADC, $ABC = 90^{0}$ and AB = BC = l. Then magnetic induction at o is

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$(\frac{1}{π} + \frac{1}{2}) \frac{μ_{o} Ι}{\sqrt{2} l}$

$(\frac{8}{π} + \frac{1}{2 \sqrt{2}}) \frac{μ_{o} Ι}{l}$

$(\frac{4}{π} + \frac{1}{\sqrt{2}}) \frac{μ_{o} Ι}{l}$

$(\frac{\sqrt{2}}{π} + \frac{1}{2 \sqrt{2}}) \frac{μ_{o} Ι}{l}$

answer is D.

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Detailed Solution

$B_{o} = 2 \times \frac{μ_{o} Ι}{4 π (\frac{l}{2})} 2 \sin 45^{0} + \frac{μ_{o} Ι}{4 (\frac{l}{\sqrt{2}})} = \frac{\sqrt{2 μ_{o}} Ι}{π l} + \frac{μ_{o} Ι}{2 \sqrt{2} l}$

$= \frac{μ_{o} Ι}{l} (\frac{\sqrt{2}}{π} + \frac{1}{2 \sqrt{2}})$

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