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Question

(A) The magnetic field due to a semi-infinite wire at a perpendicular distance $r$ is given by $B = \frac{\mu_0 i}{4 \pi r} (\sin \phi_1 + \sin \phi_2

Vedclass · Accepted Answer

$\frac{\mu_0 i}{4 \pi d}[\sqrt{2}-1]$

Vedclass · Answer

(A) The magnetic field due to a semi-infinite wire at a perpendicular distance $r$ is given by $B = \frac{\mu_0 i}{4 \pi r} (\sin \phi_1 + \sin \phi_2)$.For the horizontal segment,the point $P$ lies on its axis,so the magnetic field due to this segment is zero.For the slanted segment,the perpendicular distance from point $P$ to the wire is $r = d \sin 45^{\circ} = \frac{d}{\sqrt{2}}$.The angles subtended at $P$ by the ends of the slanted wire are $\phi_1 = 90^{\circ}$ (from the bend $Q$) and $\phi_2 = 45^{\circ}$ (since the wire is bent at $45^{\circ}$ relative to the horizontal).Using the formula $B = \frac{\mu_0 i}{4 \pi r} (\sin 90^{\circ} - \sin 45^{\circ})$ (the minus sign arises because the point $P$ is outside the projection of the segment):$B = \frac{\mu_0 i}{4 \pi (d/\sqrt{2})} (1 - \frac{1}{\sqrt{2}})$$B = \frac{\mu_0 i \sqrt{2}}{4 \pi d} (\frac{\sqrt{2}-1}{\sqrt{2}})$$B = \frac{\mu_0 i}{4 \pi d} (\sqrt{2}-1)$.

$A$ long straight wire carrying electric current '$i$' is bent at its point $Q$ to form an angle of $45^{\circ}$ as shown in the figure. The magnetic field at a point $P$ at a distance $d$ from the point $Q$ is:

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