Find the magnetic field at point P in the giv | vedclass

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Question

(C) The magnetic field due to a finite straight wire at a perpendicular distance $a$ is given by $B = \frac{{\mu _0 i}}{{4\pi a}}(\sin 	heta _1 + \si

Vedclass · Accepted Answer

$\frac{{\mu _0 i}}{{8\pi a}}(\sqrt 3 - 1) \odot $

Vedclass · Answer

(C) The magnetic field due to a finite straight wire at a perpendicular distance $a$ is given by $B = \frac{{\mu _0 i}}{{4\pi a}}(\sin 	heta _1 + \sin 	heta _2)$.Alternatively, using the angles with the wire, $B = \frac{{\mu _0 i}}{{4\pi a}}(\cos \alpha _1 - \cos \alpha _2)$, where $\alpha _1$ and $\alpha _2$ are the angles made by the position vector of point $P$ with the wire.From the diagram, the perpendicular distance from $P$ to the wire is $a$. The angles subtended by the ends of the wire at point $P$ with the perpendicular are $30^\circ$ and $60^\circ$.Thus, $\alpha _1 = 30^\circ$ and $\alpha _2 = 60^\circ$.Substituting these values: $B = \frac{{\mu _0 i}}{{4\pi a}}(\cos 30^\circ - \cos 60^\circ)$.$B = \frac{{\mu _0 i}}{{4\pi a}}\left( \frac{{\sqrt 3 }}{2} - \frac{1}{2} ight) = \frac{{\mu _0 i}}{{8\pi a}}(\sqrt 3 - 1)$.The direction is given by the right-hand rule, which is outwards $(\odot)$.

Find the magnetic field at point $P$ in the given diagram.

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