Question

Current i flows through a long conducting wire bent at right angle as shown in figure. The magnetic field at a point P on the right bisector of the angle XOY at a distance r from O is

• A. $\frac{{\mu }_{0}i}{\pi r}$
  
• B. $\frac{2{\mu }_{0}i}{\pi r}$
  
• C. $\frac{{\mu }_{0}i}{4\pi r}(\sqrt{2}+1)$
  
• D. $\frac{{\mu }_0}{4\pi }.\frac{2i}{r}(\sqrt{2}+1)$

Answer 1

The correct option is D $\frac{{\mu }_0}{4\pi }.\frac{2i}{r}(\sqrt{2}+1)$

By using $B=\frac{{\mu }_0}{4\pi }.\frac{i}{d}(sin{\varphi }_1+sin{\varphi }_2)$, from figure $d=rsin{45}^{\circ }=\frac{r}{\sqrt{2}}$
Magnetic field due to each wire at P $B=\frac{{\mu }_0}{4\pi }.\frac{i}{\left(r/\sqrt{2}\right)}(sin{45}^{\circ }+sin{90}^{\circ })$
$=\frac{{\mu }_0}{4\pi }.\frac{i}{r}(\sqrt{2}+1)$
Hence net magnetic field at P $B_{net}=2\times \frac{{\mu }_0}{4\pi }.\frac{i}{r}(\sqrt{2}+1)$