A circular arc of radius ' $\mathrm{r}$ ' carrying current ' $\mathrm{I}$ ' subtends an angle…
A circular arc of radius ' $\mathrm{r}$ ' carrying current ' $\mathrm{I}$ ' subtends an angle $\frac{\pi}{16}$ at its centre. The radius of a metal wire is uniform. The magnetic induction at the centre of circular arc is $\left[\mu_0=\right.$ permeability of free space]
$\frac{\mu_0 \mathrm{I}}{32 \mathrm{r}}$
$\frac{\mu_0 \mathrm{I}}{16 \mathrm{r}}$
$\frac{\mu_0 \mathrm{I}}{64 \mathrm{r}}$
$\frac{\mu_0 I}{8 \mathrm{r}}$
Solution
The magnetic field due to current carrying circular arc is $\mathrm{B}=\frac{\mu_0 \mathrm{I}}{2 \mathrm{r}}\left(\frac{\theta}{2 \pi}\right)$
Here, $\theta=\frac{\pi}{16}$
$\begin{aligned}
\therefore \quad \mathrm{B} & =\frac{\mu_0 \mathrm{I}}{2 \mathrm{r}}\left(\frac{1}{2 \pi} \times \frac{\pi}{16}\right) \\
\mathrm{B} & =\frac{\mu_0 \mathrm{I}}{2 \mathrm{r}}\left(\frac{1}{32}\right) \\
\mathrm{B} & =\frac{\mu_0 \mathrm{I}}{64 \mathrm{r}}
\end{aligned}$