Question

A long straight wire carrying current $I$ is bent at its mid-point to form an angle of ${45}^{\circ }$. The field at point $P$ as shown in the figure is

• A. $\frac{{\mu }_{0}I}{4\pi R}(\sqrt{2}-1)$
• B. $\frac{{\mu }_{0}I}{4\pi R}(\sqrt{2}+1)$
• C. $\frac{{\mu }_{0}I}{4\sqrt{2}\pi R}(\sqrt{2}+1)$
• D. $\frac{{\mu }_{0}I}{4\sqrt{2}\pi R}(\sqrt{2}-1)$

Answer 1

The correct option is A $\frac{{\mu }_{0}I}{4\pi R}(\sqrt{2}-1)$

Since the point P lies on the axis of straight part OA and hence magnetic field due to this part will be zero.

as

$B=\frac{{\mu }_{0}I}{4\pi r}(sin\alpha +sin\beta )$

for this case

Since both ends O and B are on the same side of normal Pn,

therefore

${\varphi }_1=-{45}^{\circ }$ and ${\varphi }_2={90}^{\circ }$

$\therefore B=\frac{{\mu }_{0}I}{4\pi d}\left(sin\right(-{45}^{\circ })+sin({90}^{\circ }\left)\right)$

$=\frac{{\mu }_{0}I}{4\pi \left(Rcos{45}^{\circ }\right)}[-sin{45}^{\circ }+1]$

$=\frac{{\mu }_{0}I}{4\pi R\ast \frac{1}{\sqrt{2}}}(1-\frac{1}{\sqrt{2}})$

$=\frac{{\mu }_{0}I}{4\pi R}(\sqrt{2}-1)$