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An infinitely long wire is bent in the form of a semicircle at the end as shown in the figure. It carries current I along abcdo. If radius of the semicircle be R, then the magnetic field at ‘O’ which is the centre of the circular part is

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$\frac{μ_{0}}{4 π} \frac{2 I}{R} (π + 1)$

$\frac{μ_{0}}{4 π} \frac{2 I}{R} (π - 1)$

$\frac{μ_{0}}{4 π} \frac{I}{R} (π + 1)$

$\frac{μ_{0}}{4 π} \frac{I}{R} (π - 1)$

answer is C.

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Detailed Solution

$B = \frac{μ_{0} I}{4 πR} + \frac{μ_{0} I}{2 R} (\frac{π}{2 π}); B = \frac{μ_{0} I}{4 πR} (1 + π)$

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