A long straight wire, carrying current I, is | vedclass

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Question

since point $P$ lies on axis of straight part ab, therefore, magnetic induction due to this part is equal to zero.

For part $bc:$ From fig

$r=R

Vedclass · Accepted Answer

$\frac{{\left( {\sqrt 2 \, - \,1} \right){\mu _0}I}}{{4\pi R}}$

Vedclass · Answer

since point $P$ lies on axis of straight part ab, therefore, magnetic induction due to this part is equal to zero.

For part $bc:$ From fig

$r=R \cos 45^{\circ}$

since both the ends $b$ and $c$ are on the same side of normal $\mathrm{PN}$, therefore, $\alpha$ is negative and $\beta$ is positive. Hence, $\alpha=-45^{\circ}$ and $\beta=+90^{\circ}$

Using $B=\frac{\mu_{0} I}{4 \pi r}(\sin \alpha+\sin \beta),$ we have $B=\frac{(\sqrt{2}-1) \mu_{0} I}{4 \pi R}$

(into the plane of paper)

A long straight wire, carrying current $I,$ is bent at its midpoint to from an angle of $45^o.$ Induction of magnetic field at point $P,$ distant $R$ from point of bending is equal to :

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