MHT CET · Physics · Electromagnetic Induction
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 \(\frac{\mu_0 \mathrm{I}}{16 \mathrm{r}}\)
- C \(\frac{\mu_0 \mathrm{I}}{64 \mathrm{r}}\)
- D \(\frac{\mu_0 I}{8 \mathrm{r}}\)
Answer & Solution
Correct Answer
(C) \(\frac{\mu_0 \mathrm{I}}{64 \mathrm{r}}\)
Step-by-step Solution
Detailed explanation
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, \(\theta=\frac{\pi}{16}\)
\(\begin{aligned}
\therefore \quad \mathrm{B} & =\frac{\mu_0 \mathrm{I}}{2 \mathrm{r}}\left(\frac{1}{2 \pi} \times \frac{\pi}{16}\right) \\
\mathrm{B} & =\frac{\mu_0 \mathrm{I}}{2 \mathrm{r}}\left(\frac{1}{32}\right) \\
\mathrm{B} & =\frac{\mu_0 \mathrm{I}}{64 \mathrm{r}}
\end{aligned}\)
Here, \(\theta=\frac{\pi}{16}\)
\(\begin{aligned}
\therefore \quad \mathrm{B} & =\frac{\mu_0 \mathrm{I}}{2 \mathrm{r}}\left(\frac{1}{2 \pi} \times \frac{\pi}{16}\right) \\
\mathrm{B} & =\frac{\mu_0 \mathrm{I}}{2 \mathrm{r}}\left(\frac{1}{32}\right) \\
\mathrm{B} & =\frac{\mu_0 \mathrm{I}}{64 \mathrm{r}}
\end{aligned}\)
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