A straight wire of finite length carrying current $l$ subtends an angle of $60^{\circ}$ at point $P$ as shown. The magnetic field at $P$ is
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(a)
$B=\frac{\mu_0 I}{4 \pi x \cos 30^{\circ}}\left[\sin 30^{\circ}+\sin 30^{\circ}\right]$
$B=\frac{\mu_0 I}{4 \pi x\left(\frac{\sqrt{3}}{2}\right)}\left(\frac{1}{2}+\frac{1}{2}\right)$
$=\frac{\mu_0 i}{2 \sqrt{3} \pi x}$
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