An infinitely long straight conductor is bent into shape as shown in figure. It carries a current $I…

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

1. $\frac{\mu_0 /(\pi-1)}{2 \pi r}$
2. $\frac{\mu_0 /(\pi+1)}{2 \pi r}$
3. $\frac{\mu_0 /(2 \pi-1)}{2 \pi r}$
4. $\frac{\mu_0 /(2 \pi+1)}{2 \pi r}$

Solution

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]

Asked in: AP EAMCET 2020 (22 Sep Shift 2)

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