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An infinitely long straight conductor is bent into shape as shown in figure. It carries a current $I \mathrm{~A}$ and the radius of circular loop is $r \mathrm{~m}$. The magnetic induction at the centre of the circular loop is
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Verified Answer
The correct answer is:
$\frac{\mu_0 /(\pi-1)}{2 \pi r}$
The given situation is shown in the figure,
Net magnetic field at $O$,
$B=B$ due to straight wire $P Q$ (upward) $+B$ due to straight wire $C D$ (upward) $+B$ due to circular wire (downward)
$$
\begin{aligned}
& =-\frac{\mu_0}{4 \pi} \cdot \frac{I}{r}-\frac{\mu_0}{4 \pi} \cdot \frac{I}{r}+\frac{\mu_0 I}{2 r} \\
& =-\frac{\mu_0 I}{2 \pi r}+\frac{\mu_0 I}{2 r}=\frac{\mu_0 I}{2 r}\left[-\frac{1}{\pi}+1\right] \\
& =\frac{\mu_0 I}{2 r}\left[\frac{-1+\pi}{\pi}\right]=\frac{\mu_0 I(\pi-1)}{2 \pi r}
\end{aligned}
$$
[downward]
Net magnetic field at $O$,
$B=B$ due to straight wire $P Q$ (upward) $+B$ due to straight wire $C D$ (upward) $+B$ due to circular wire (downward)
$$
\begin{aligned}
& =-\frac{\mu_0}{4 \pi} \cdot \frac{I}{r}-\frac{\mu_0}{4 \pi} \cdot \frac{I}{r}+\frac{\mu_0 I}{2 r} \\
& =-\frac{\mu_0 I}{2 \pi r}+\frac{\mu_0 I}{2 r}=\frac{\mu_0 I}{2 r}\left[-\frac{1}{\pi}+1\right] \\
& =\frac{\mu_0 I}{2 r}\left[\frac{-1+\pi}{\pi}\right]=\frac{\mu_0 I(\pi-1)}{2 \pi r}
\end{aligned}
$$
[downward]
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