The magnitude of magnetic field at point ' $O$ ' in the following figure will be

The magnitude of magnetic field at point ' $O$ ' in the following figure will be
  1. $\frac{\mu_0}{4 \pi} \frac{\mathrm{I}}{\mathrm{r}}\left(\frac{2}{\pi}+2\right)$
  2. $\frac{\mu_0}{4 \pi} \frac{\mathrm{I}}{\mathrm{r}}\left(\frac{2}{\pi}-2\right)$
  3. $\frac{\mu_0}{4 \pi} \frac{\mathrm{I}}{\mathrm{r}}\left(2+\frac{\pi}{2}\right)$
  4. $\frac{\mu_0}{4 \pi} \frac{\mathrm{I}}{\mathrm{r}}\left(2-\frac{\pi}{2}\right)$

Solution

Magnetic field due to current carrying arc is, $B=\frac{\mu_0 I}{4 \pi r} \times \theta=\frac{\mu_0 I}{4 \pi r} \times \frac{\pi}{2}=\frac{\mu_0 I}{8 r}$
Magnetic field due to semi-infinite current carrying straight wires $A B$ and $C D$ is, $\frac{2 \mu_0 I}{4 \pi r}=\frac{\mu_0 I}{2 \pi r}$ $\therefore \quad$ Total magnetic field at ' O ' will be, $\begin{aligned} \frac{\mu_0 I}{2 \pi r}+\frac{\mu_0 I}{8 r} & =\frac{\mu_0 I}{2 r}\left(\frac{1}{\pi}+\frac{1}{4}\right) \\ & =\frac{\mu_0}{4 \pi} \frac{1}{r}\left(\frac{2 \pi}{\pi}+\frac{2 \pi}{4}\right) \\ & =\frac{\mu_0}{4 \pi} \frac{1}{r}\left(2+\frac{\pi}{2}\right) \end{aligned}$ ^

Asked in: MHT CET 2024 (10 May Shift 2)

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