A long straight wire, carrying current I, is | vedclass

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(B) Let the wire be bent at point $b.$ The wire consists of two semi-infinite segments. Point $P$ lies on the axis of the first segment, so the magnet

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$\frac{(\sqrt{2}+1)\mu_{0}I}{4\pi R}$

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(B) Let the wire be bent at point $b.$ The wire consists of two semi-infinite segments. Point $P$ lies on the axis of the first segment, so the magnetic field due to this segment at $P$ is $0.$For the second segment, let the perpendicular distance from $P$ to the line of the wire be $r.$From the geometry, $r = R \sin 45^{\circ} = \frac{R}{\sqrt{2}}.$The magnetic field due to a semi-infinite wire at a perpendicular distance $r$ is $B = \frac{\mu_{0}I}{4\pi r}(1 + \sin 	heta),$ where $	heta$ is the angle subtended by the end of the wire at $P.$ Here, one end is at infinity $(	heta = 90^{\circ})$ and the other end is at the bend $(b).$ The angle subtended by the bend at $P$ relative to the normal is $45^{\circ}.$Thus, $B = \frac{\mu_{0}I}{4\pi (R/\sqrt{2})} (\sin 90^{\circ} + \sin 45^{\circ}) = \frac{\sqrt{2}\mu_{0}I}{4\pi R} (1 + \frac{1}{\sqrt{2}}) = \frac{\mu_{0}I}{4\pi R} (\sqrt{2} + 1).$

$A$ long straight wire, carrying current $I,$ is bent at its midpoint to form an angle of $45^{\circ}.$ The magnetic field induction at point $P,$ at a distance $R$ from the point of bending, is equal to:

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