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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. Magnetic field at a point $P$ at a distance $d$ from the point $Q$ of bending is
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Verified Answer
The correct answer is:
$\frac{\mu_0 i}{4 \pi d}[\sqrt{2}-1]$
Magnetic field at a point from a current carrying conductor making angle $\theta_1$ and $\theta_2$ with the ends of wire is given using Bio-Savart's law as
$$
\begin{aligned}
B & =\frac{\mu_0 I}{4 \pi r}\left(\sin \theta_1+\sin \theta_2\right) \\
& =\frac{\mu_0 I}{4 \pi\left(\frac{d}{\sqrt{2}}\right)}\left[\sin 90^{\circ}+\sin 135^{\circ}\right] \\
& =\frac{\sqrt{2} \mu_0 I}{4 \pi d}\left(1-\frac{1}{\sqrt{2}}\right)=\frac{\mu_0 I}{4 \pi d}(\sqrt{2}-1)
\end{aligned}
$$
$$
\begin{aligned}
B & =\frac{\mu_0 I}{4 \pi r}\left(\sin \theta_1+\sin \theta_2\right) \\
& =\frac{\mu_0 I}{4 \pi\left(\frac{d}{\sqrt{2}}\right)}\left[\sin 90^{\circ}+\sin 135^{\circ}\right] \\
& =\frac{\sqrt{2} \mu_0 I}{4 \pi d}\left(1-\frac{1}{\sqrt{2}}\right)=\frac{\mu_0 I}{4 \pi d}(\sqrt{2}-1)
\end{aligned}
$$
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