Q123 / 150 · MHT CET 2023 (11 May Shift 2) · Physics

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]

  • A
    $\frac{\mu_0 \mathrm{I}}{32 \mathrm{r}}$
  • B
    $\frac{\mu_0 \mathrm{I}}{16 \mathrm{r}}$
  • Correct answer
    C
    $\frac{\mu_0 \mathrm{I}}{64 \mathrm{r}}$
  • D
    $\frac{\mu_0 I}{8 \mathrm{r}}$

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, $\theta=\frac{\pi}{16}$ $\begin{aligned} \therefore \quad \mathrm{B} & =\frac{\mu_0 \mathrm{I}}{2 \mathrm{r}}\left(\frac{1}{2 \pi} \times \frac{\pi}{16}\right) \\ \mathrm{B} & =\frac{\mu_0 \mathrm{I}}{2 \mathrm{r}}\left(\frac{1}{32}\right) \\ \mathrm{B} & =\frac{\mu_0 \mathrm{I}}{64 \mathrm{r}} \end{aligned}$

Hence, the correct answer is option: 'C'.

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