A long conducting wire carrying a current I i | vedclass

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(D) The magnetic field due to a finite straight wire at a perpendicular distance $r$ is given by $B = \frac{\mu_{0} I}{4 \pi r} (\sin 	heta_{1} + \si

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$\frac{\sqrt{3} \mu_{0} I}{2 \pi d}$

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(D) The magnetic field due to a finite straight wire at a perpendicular distance $r$ is given by $B = \frac{\mu_{0} I}{4 \pi r} (\sin 	heta_{1} + \sin 	heta_{2})$.For the given geometry, the point $P$ is at a distance $d$ from the bend along the angle bisector. The perpendicular distance from $P$ to each wire segment is $r = d \sin(60^{\circ}) = \frac{d \sqrt{3}}{2}$.The angles subtended by the ends of the wire segments at point $P$ are $	heta_{1} = 90^{\circ}$ and $	heta_{2} = 30^{\circ}$.Since there are two identical segments, the total magnetic field is $B_{	ext{net}} = 2 	imes \frac{\mu_{0} I}{4 \pi r} (\sin 90^{\circ} + \sin 30^{\circ})$.Substituting the values:$B_{	ext{net}} = 2 	imes \frac{\mu_{0} I}{4 \pi (d \sqrt{3} / 2)} 	imes (1 + 1/2)$$B_{	ext{net}} = 2 	imes \frac{\mu_{0} I}{2 \pi d \sqrt{3}} 	imes \frac{3}{2}$$B_{	ext{net}} = \frac{3 \mu_{0} I}{2 \pi d \sqrt{3}} = \frac{\sqrt{3} \mu_{0} I}{2 \pi d}$.

$A$ long conducting wire carrying a current $I$ is bent at $120^{\circ}$ (see figure). The magnetic field $B$ at a point $P$ on the angle bisector of the bend at a distance $d$ from the bend is ($\mu_{0}$ is the permeability of free space):

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