An arc of a circle of radius ' $R$ ' subtends an angle $\frac{\pi}{2}$ at the centre. It carries a current…

An arc of a circle of radius ' $R$ ' subtends an angle $\frac{\pi}{2}$ at the centre. It carries a current $I$. The magnetic field at the centre will be ( $\mu_0=$ permeability of free space)
  1. $\frac{\mu_0 I}{2 R}$
  2. $\frac{\mu_0 I}{8 R}$
  3. $\frac{\mu_0 I}{4 R}$
  4. $\frac{2 \mu_0 I}{5 R}$

Solution

The magnetic field due to current carrying circular arc is $B=\frac{\mu_0 I}{2 r}\left(\frac{\theta}{2 \pi}\right)$
Here, $\theta=\frac{\pi}{2}$ $\begin{aligned} \therefore \quad B & =\frac{\mu_0 I}{2 r}\left(\frac{1}{2 \pi} \times \frac{\pi}{2}\right) \\ B & =\frac{\mu_0 I}{2 r}\left(\frac{1}{4}\right) \\ B & =\frac{\mu_0 I}{8 r} \end{aligned}$ :

Asked in: MHT CET 2024 (16 May Shift 2)

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