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Question

(A) The magnetic field $B$ at a distance $r$ from a finite wire carrying current $i$ is given by $B = \frac{\mu_0 i}{4 \pi r} (\sin 	heta_1 + \sin

Vedclass · Accepted Answer

$\frac{{\mu _0}i}{{4\pi d}}\left( {\sqrt 2 - 1} \right)$

Vedclass · Answer

(A) The magnetic field $B$ at a distance $r$ from a finite wire carrying current $i$ is given by $B = \frac{\mu_0 i}{4 \pi r} (\sin 	heta_1 + \sin 	heta_2)$.In the given figure,the perpendicular distance from point $p$ to the wire is $r = d \cos(45^{\circ}) = \frac{d}{\sqrt{2}}$.The angles subtended by the ends of the wire at point $p$ are $	heta_1 = 90^{\circ}$ (for the infinite end) and $	heta_2 = -45^{\circ}$ (for the finite end).Substituting these values into the formula:$B = \frac{\mu_0 i}{4 \pi (d / \sqrt{2})} (\sin 90^{\circ} + \sin(-45^{\circ}))$$B = \frac{\sqrt{2} \mu_0 i}{4 \pi d} (1 - \frac{1}{\sqrt{2}})$$B = \frac{\mu_0 i}{4 \pi d} (\sqrt{2} - 1)$.

Find the magnitude of the magnetic field at point $p$ due to the semi-infinite wire shown below.

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