What is the magnetic field at point P in the | vedclass

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(D) The magnetic field due to a finite straight wire at a point $P$ is given by the formula: $B = \frac{{{\mu _0}i}}{{4\pi r}}(\sin {\phi _1} + \sin {

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$\frac{{\sqrt 2 {\mu _0}i}}{{8\pi l}}$

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(D) The magnetic field due to a finite straight wire at a point $P$ is given by the formula: $B = \frac{{{\mu _0}i}}{{4\pi r}}(\sin {\phi _1} + \sin {\phi _2})$ where $r$ is the perpendicular distance from the wire to point $P$. From the figure,the perpendicular distance $r = l$. The wire extends from the level of point $P$ downwards to a distance $l$. Thus,the angles subtended at point $P$ by the ends of the wire are $\phi _1 = 0^\circ$ and $\phi _2 = 45^\circ$ (since the wire length is $l$ and the horizontal distance is $l$,$	an \phi _2 = l/l = 1$,so $\phi _2 = 45^\circ$). Substituting these values into the formula: $B = \frac{{{\mu _0}i}}{{4\pi l}}(\sin 0^\circ + \sin 45^\circ)$ $B = \frac{{{\mu _0}i}}{{4\pi l}}(0 + \frac{1}{{\sqrt 2 }})$ $B = \frac{{{\mu _0}i}}{{4\sqrt 2 \pi l}} 	imes \frac{{\sqrt 2 }}{{\sqrt 2 }} = \frac{{\sqrt 2 {\mu _0}i}}{{8\pi l}}$

What is the magnetic field at point $P$ in the given figure?

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