An infinitely long straight conductor is bent | vedclass

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Question

(b) The given shape is equivalent to the following diagram
The field at $O$ due to straight part of conductor is ${B_1} = \frac{{{\mu _o}}}{{4\pi }}.

Vedclass · Accepted Answer

$\frac{{{\mu _0}}}{{4\pi }}\frac{{2i}}{r}(\pi - 1)$

Vedclass · Answer

(b) The given shape is equivalent to the following diagram
The field at $O$ due to straight part of conductor is ${B_1} = \frac{{{\mu _o}}}{{4\pi }}.\frac{{2i}}{r}  \odot $. The field at $O$ due to circular coil is ${B_2} = \frac{{{\mu _0}}}{{4\pi }}.\frac{{2\pi i}}{r} \otimes $. Both fields will act in the opposite direction, hence the total field at $O$.
i.e. $B = {B_2} - {B_1} = \left( {\frac{{{\mu _o}}}{{4\pi }}} ight) 	imes (\pi - 1)\frac{{2i}}{r} = \frac{{{\mu _o}}}{{4\pi }}.\frac{{2i}}{r}(\pi - 1)$

An infinitely long straight conductor is bent into the shape as shown in the figure. It carries a current of $i$ $ampere$ and the radius of the circular loop is $r$ $metre$. Then the magnetic induction at its centre will be

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