Question

The magnetic field at the centre O of the arc shown in the figure is

• A. $2I\left(\sqrt{2}+\pi \right)\times \frac{10^{-7}}{r}$
• B. $2I\left(\sqrt{2}+\frac{\pi }{4}\right)\times \frac{10^{-7}}{r}$
• C. $I\left(\sqrt{2}+\pi \right)\times \frac{10^{-7}}{r}$
• D. $I\left(\sqrt{2}+\frac{\pi }{4}\right)\times \frac{10^{-7}}{r}$

Answer 1

Answer: (B)

Here, $a=\frac{r}{\sqrt{2}}$
Magnetic field at point O due to AB is
$B_1=\frac{{\mu }_0}{4\pi }\frac{I}{a}=\frac{{\mu }_{0}I}{4\pi \left(r/\sqrt{2}\right)}$
Magnetic field at point O due to BCD is
$B_2=\frac{{\mu }_{0}I}{4\pi r}\left(\frac{\pi }{2}\right)$
Magnetic field at point O due to DE is
$B_3=\frac{{\mu }_{0}I}{4\pi a}=\frac{{\mu }_{0}I}{4\pi \left(r/\sqrt{2}\right)}$
Resultant magnetic field at point O is
$B=B_1+B_2+B_3$
$=\frac{{\mu }_{0}I}{4\pi \left(r/\sqrt{2}\right)}+\frac{{\mu }_{0}I}{4\pi r}\left(\frac{\pi }{2}\right)+\frac{{\mu }_{0}I}{4\pi \left(r/\sqrt{2}\right)}$
$=\frac{{\mu }_{0}I}{4\pi r}\left(\sqrt{2}+\frac{\pi }{2}+\sqrt{2}\right)=\frac{{\mu }_{0}I}{4\pi r}\left(2\sqrt{2}+\frac{\pi }{2}\right)$
$=\frac{{\mu }_{0}2I}{4\pi r}\left(\sqrt{2}+\frac{\pi }{4}\right)$
$=\frac{10^{-7}\times 2I}{r}\left(\sqrt{2}+\frac{\pi }{4}\right)\left[\because \frac{{\mu }_0}{4\pi }=10^{-7}\right]$