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Question

(A) The magnetic field at point $P$ due to the straight wire segment carrying current $i$ that passes through $P$ is zero,because point $P$ lies on th

Vedclass · Accepted Answer

$\frac{\mu_0 i}{2 \sqrt{2} \pi d}$

Vedclass · Answer

(A) The magnetic field at point $P$ due to the straight wire segment carrying current $i$ that passes through $P$ is zero,because point $P$ lies on the axis of this segment.For the second segment,the perpendicular distance $r$ from point $P$ to the line of the wire is $r = d \sin(45^{\circ}) = \frac{d}{\sqrt{2}}$.The magnetic field $B$ due to a semi-infinite wire at a perpendicular distance $r$ is given by $B = \frac{\mu_0 i}{4 \pi r}$.Substituting $r = \frac{d}{\sqrt{2}}$,we get $B = \frac{\mu_0 i}{4 \pi (d / \sqrt{2})} = \frac{\sqrt{2} \mu_0 i}{4 \pi d} = \frac{\mu_0 i}{2 \sqrt{2} \pi d}$.

$A$ long straight wire carrying electric current $i$ is bent at its mid-point 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 $Q$ of bending is:

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