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A hairpin like shape as shown in figure is made by bending a long current carrying wire. What is the magnitude of a magnetic field at point P which lies on the centre of the semicircle ?

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$\frac{μ_{0} I}{4π r} (2 - π)$

$\frac{μ_{0} I}{4π r} (2 + π)$

$\frac{μ_{0} I}{2π r} (2 - π)$

$\frac{μ_{0} I}{2π r} (2 + π)$

answer is A.

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Detailed Solution

Magnetic field at P, B_P = 2 (Magnetic field due to straight wire) + (Magnetic field due to semicircle)

$B_{P} = 2 (\frac{μ_{0} l}{4 πr}) + \frac{μ_{0} I}{4 r}$

$∴ B_{P} = \frac{μ_{0} I}{2πr} + \frac{μ_{0} I}{4r} = \frac{μ_{0} I}{4πr} (2+π)$

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