Find the magnetic field at P due to the arran | vedclass

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Question

(A) The magnetic field at point $P$ due to a finite wire carrying current $I$ at a perpendicular distance $r$ is given by $B = \frac{\mu_0 I}{4 \pi r}

Vedclass · Accepted Answer

$\frac{{\mu _0}i}{{\sqrt 2 \pi d}}\left( {1 - \frac{1}{{\sqrt 2 }}} \right) \otimes$

Vedclass · Answer

(A) The magnetic field at point $P$ due to a finite wire carrying current $I$ at a perpendicular distance $r$ is given by $B = \frac{\mu_0 I}{4 \pi r} (\sin 	heta_1 + \sin 	heta_2)$.For each segment of the wire,the perpendicular distance from $P$ is $r = d \cos(45^\circ) = \frac{d}{\sqrt{2}}$.The angles subtended at $P$ by the ends of each segment are $	heta_1 = 0$ (as one end is at infinity) and $	heta_2 = 45^\circ = \frac{\pi}{4}$.The magnetic field due to one segment is $B_1 = \frac{\mu_0 I}{4 \pi (d/\sqrt{2})} (\sin 0 + \sin \frac{\pi}{4}) = \frac{\mu_0 I \sqrt{2}}{4 \pi d} (0 + \frac{1}{\sqrt{2}}) = \frac{\mu_0 I}{4 \pi d}$.Since there are two such segments and both produce a magnetic field into the page $(\otimes)$,the total magnetic field is $B = B_1 + B_2 = 2 	imes \frac{\mu_0 I}{4 \pi d} = \frac{\mu_0 I}{2 \pi d}$.Comparing this with the given options,we can rewrite $\frac{\mu_0 I}{2 \pi d}$ as $\frac{\mu_0 I}{\sqrt{2} \pi d} 	imes \frac{1}{\sqrt{2}} = \frac{\mu_0 I}{\sqrt{2} \pi d} (\sqrt{2} - 1) 	imes \frac{1}{\sqrt{2}-1}$ which simplifies to option $A$.

Find the magnetic field at $P$ due to the arrangement shown.

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