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Physics Question on Magnetic Field

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

A

2I(2+π)×10−7r2I\left(\sqrt{2}+\pi\right)\times\frac{10^{-7}}{r}

B

2I(2+π4)×10−7r2I\left(\sqrt{2}+\frac{\pi}{4}\right)\times\frac{10^{-7}}{r}

C

I(2+π)×10−7rI\left(\sqrt{2}+\pi\right)\times\frac{10^{-7}}{r}

D

I(2+π4)×10−7rI\left(\sqrt{2}+\frac{\pi}{4}\right)\times\frac{10^{-7}}{r}

Answer

2I(2+π4)×10−7r2I\left(\sqrt{2}+\frac{\pi}{4}\right)\times\frac{10^{-7}}{r}

Explanation

Solution

Here, a=r2a =\frac{r}{\sqrt{2}} Magnetic field at point O due to AB is B1=μ04πIa=μ0I4π(r/2)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 B2=μ0I4πr(π2)B_{2}=\frac{\mu_{0}I}{4\pi r} \left(\frac{\pi}{2}\right) Magnetic field at point O due to DE is B3=μ0I4πa=μ0I4π(r/2)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=B1+B2+B3B = B_{1} + B_{2} + B_{3} =μ0I4π(r/2)+μ0I4πr(π2)+μ0I4π(r/2)=\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)} =μ0I4πr(2+π2+2)=μ0I4πr(22+π2)=\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) =μ02I4πr(2+π4)=\frac{\mu_{0}2I}{4\pi r}\left(\sqrt{2}+\frac{\pi}{4}\right) =10−7×2Ir(2+π4)[∵μ04π=10−7]=\frac{10^{-7}\times2I}{r}\left(\sqrt{2}+\frac{\pi}{4}\right)\quad\quad\quad\left[\because \frac{\mu_{0}}{4\pi}=10^{-7}\right]