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Question

(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

Vedclass · Accepted Answer

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

Vedclass · Answer

(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}$.

$(i)$	$(ii)$	$(iii)$
$(A). \frac{\mu_0 i}{2r} \odot$	$(A). \frac{\mu_0}{2\pi} \frac{i}{r}(\pi - 2)$	$(A). \frac{\mu_0}{2r} \frac{2i}{r}(\pi + 1) \otimes$
$(B). \frac{\mu_0 i}{2r} \otimes$	$(B). \frac{\mu_0 i}{4\pi} \frac{i}{r}(\pi + 2) \otimes$	$(B). \frac{\mu_0 i}{4r} \frac{2i}{r}(\pi - 1) \otimes$
$(C). \frac{3\mu_0 i}{8r} \otimes$	$(C). \frac{\mu_0 i}{4r} \otimes$	$(C). \text{Zero}$
$(D). \frac{3\mu_0 i}{8r} \odot$	$(D). \frac{\mu_0 i}{4r} \odot$	$(D). \text{Infinite}$

$A$ 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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