Key idea

The loop consists of:

• two circular arcs with common center $O$
• two radial segments joining them

We need the magnetic field at $O$.

Contribution of radial segments

For a radial wire segment, the current element $d\vec{\ell}$ is along the radius vector from the segment to $O$. Hence for every element, $d\vec{B} \propto d\vec{\ell} \times \hat{r} = 0$ So, both radial lines produce zero magnetic field at $O$.

Field due to a circular arc

Magnetic field at the center due to an arc of angle $\theta$ is $B = \frac{\mu_0 I\theta}{4\pi R}$

From the figure, the arcs subtend $270^\circ = \frac{3\pi}{2}$ at the center.

Therefore, for the two arcs:

• inner arc radius $r_1 = 10\,\text{cm} = 0.1\,\text{m}$
• outer arc radius $r_2 = 20\,\text{cm} = 0.2\,\text{m}$
• current $I = 10\,\text{A}$

Since current flows along the two arcs in opposite senses around $O$, their magnetic fields at $O$ oppose each other.

So net field is $B_{\text{net}} = \frac{\mu_0 I\theta}{4\pi}\left(\frac{1}{r_1} - \frac{1}{r_2}\right)$

Substitute values

Using $\mu_0 = 4\pi \times 10^{-7}\,\text{T m/A}, \quad \theta = \frac{3\pi}{2}$

we get $B_{\text{net}} = \frac{(4\pi \times 10^{-7})(10)}{4\pi}\cdot \frac{3\pi}{2} \left(\frac{1}{0.1} - \frac{1}{0.2}\right)$

Simplify: $= 10^{-6} \cdot \frac{3\pi}{2} \cdot (10 - 5)$ $= 10^{-6} \cdot \frac{3\pi}{2} \cdot 5$ $= 7.5\pi \times 10^{-6}$

$B_{\text{net}} \approx 23.6 \times 10^{-6}\,\text{T} = 2.36 \times 10^{-5}\,\text{T}$

Choose the closest option

The nearest value is $1.5 \times 10^{-5}\,\text{T}$

So the correct option is C.

Comparison with stored answer

Stored correct answer: D ($1.0 \times 10^{-5}\,\text{T}$)

My derived answer is C ($1.5 \times 10^{-5}\,\text{T}$), because the standard arc-field formula gives a value closer to $1.5\times10^{-5}$ than to $1.0\times10^{-5}$.

Hence I disagree with the stored answer.