As shown in the figure, a long straight conductor with a semicircular arc of radius $$\frac{\pi}{10}$$ m is carrying a current $$I = 3A$$. The magnitude of the magnetic field at the center $$O$$ of the arc is: (The permeability of the vacuum $$= 4\pi \times 10^{-7}$$ NA$$^{-2}$$)

Answer in $$\mu$$T.

For a circular arc, the field at the center is $$B = \frac{\mu_0 I}{4\pi R} \theta$$, where $$\theta$$ is the angle in radians. A semicircle ($$\theta = \pi$$) produces $$B = \frac{\mu_0 I}{4R}$$

For the straight parts of the wire, the point $$O$$ lies exactly on the line of the conductor. This means the angle ($$\theta$$) between the current element $$d\vec{l}$$ and the position vector $$\vec{r}$$ is $$0^\circ$$ or $$180^\circ$$. Since $$\sin(0^\circ) = \sin(180^\circ) = 0$$, the magnetic field contribution from both straight sections is zero.

$$B = \frac{(4\pi \times 10^{-7}) \times 3}{4 \times \left(\frac{\pi}{10}\right)}$$

$$B = \frac{12\pi \times 10^{-7}}{\frac{4\pi}{10}} = \frac{12\pi \times 10^{-7} \times 10}{4\pi}$$

$$B = 3 \times 10 \times 10^{-7} = 3 \times 10^{-6}\text{ T}$$ 

$$\mathbf{B = 3\text{ }\mu\text{T}}$$