Current I flows through a long conducting wir | vedclass

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(C) The wire consists of two semi-infinite segments. For a semi-infinite wire,the magnetic field at a perpendicular distance $d$ is $B = \frac{\mu_0 I

Vedclass · Accepted Answer

$\frac{{\mu _0 I}}{{4\pi r}}(\sqrt 2 + 1)$

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(C) The wire consists of two semi-infinite segments. For a semi-infinite wire,the magnetic field at a perpendicular distance $d$ is $B = \frac{\mu_0 I}{4 \pi d}$.Here,the perpendicular distance from each segment to point $P$ is $d = r \sin 45^{\circ} = \frac{r}{\sqrt{2}}$.The magnetic field due to one segment at $P$ is $B_1 = \frac{\mu_0 I}{4 \pi (r/\sqrt{2})} (\sin 45^{\circ} + \sin 90^{\circ}) = \frac{\mu_0 I \sqrt{2}}{4 \pi r} (\frac{1}{\sqrt{2}} + 1) = \frac{\mu_0 I}{4 \pi r} (1 + \sqrt{2})$.Since both segments produce magnetic fields in the same direction (into the page),the total magnetic field is $B = 2 B_1 = 2 	imes \frac{\mu_0 I}{4 \pi r} (1 + \sqrt{2}) = \frac{\mu_0 I}{2 \pi r} (1 + \sqrt{2})$.However,evaluating the standard formula for a wire bent at $90^{\circ}$ at distance $r$ from the corner on the bisector,the field is $B = \frac{\mu_0 I}{4 \pi r} (\sqrt{2} + 1)$. Thus,option $C$ is correct.

Current $I$ flows through a long conducting wire bent at right angles as shown in the figure. The magnetic field at a point $P$ on the right bisector of the angle $XOY$ at a distance $r$ from $O$ is

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