An infinitely long straight conductor is bent | vedclass

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Question

(B) The magnetic field at the center $O$ is due to two parts: the straight wire and the circular loop.$1$. The magnetic field due to the circular loop

Vedclass · Accepted Answer

$\frac{\mu_0}{4\pi} \frac{2i}{r}(\pi - 1)$

Vedclass · Answer

(B) The magnetic field at the center $O$ is due to two parts: the straight wire and the circular loop.$1$. The magnetic field due to the circular loop of radius $r$ carrying current $i$ is $B_{loop} = \frac{\mu_0 i}{2r} = \frac{\mu_0}{4\pi} \frac{2\pi i}{r}$,directed into the plane of the paper (using the right-hand rule).$2$. The magnetic field due to the straight wire at a distance $r$ from it is $B_{wire} = \frac{\mu_0 i}{2\pi r} = \frac{\mu_0}{4\pi} \frac{2i}{r}$,directed out of the plane of the paper.$3$. Since the fields are in opposite directions,the net magnetic field $B$ at $O$ is $B = B_{loop} - B_{wire} = \frac{\mu_0}{4\pi} \frac{2\pi i}{r} - \frac{\mu_0}{4\pi} \frac{2i}{r} = \frac{\mu_0}{4\pi} \frac{2i}{r}(\pi - 1)$.

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 $O$ will be:

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