The magnetic field at the centre O of the arc | vedclass

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Question

(b)

Here, $a=\frac{r}{\sqrt{2}}$

Magnetic field at point $O$ due to $A B$ is

$B_1=\frac{\mu_0}{4 \pi} \frac{I}{a}=\frac{\mu_0 I}{4 \pi(r / \s

Vedclass · Accepted Answer

$2 I (\sqrt{2}+\pi) 	imes \frac{10^{-7}}{ r }$

Vedclass · Answer

(b)

Here, $a=\frac{r}{\sqrt{2}}$

Magnetic field at point $O$ due to $A B$ is

$B_1=\frac{\mu_0}{4 \pi} \frac{I}{a}=\frac{\mu_0 I}{4 \pi(r / \sqrt{2})}$

Magnetic field at point $O$ due to $BCD$ is

$B_2=\frac{\mu_0 I}{4 \pi r}\left(\frac{\pi}{2}ight)$

Magnetic field at point $O$ due to $D E$ is

$B_3=\frac{\mu_0 I}{4 \pi a}=\frac{\mu_0 I}{4 \pi(r / \sqrt{2})}$

Resultant magnetic field at point $O$ is

$B=B_1+B_2+B_3$

$=\frac{\mu_0 I}{4 \pi(r / \sqrt{2})}+\frac{\mu_0 I}{4 \pi r}\left(\frac{\pi}{2}ight)+\frac{\mu_0 I}{4 \pi(r / \sqrt{2})}$

$=\frac{\mu_0 I}{4 \pi r}\left(\sqrt{2}+\frac{\pi}{2}+\sqrt{2}ight)=\frac{\mu_0 I}{4 \pi r}\left(2 \sqrt{2}+\frac{\pi}{2}ight)$

$=\frac{\mu_0 2 I}{4 \pi r}\left(\sqrt{2}+\frac{\pi}{4}ight)$

$=\frac{10^{-7} 	imes 2 I}{r}\left(\sqrt{2}+\frac{\pi}{4}ight) \quad\left[\because \frac{\mu_0}{4 \pi}=10^{-7}ight]$

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

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