1. Identify which parts of the loop produce magnetic field at the origin

The loop consists of:

• inner arc $DA$ of radius $a$,
• outer arc $BC$ of radius $b$,
• two straight radial segments $AB$ and $CD$.

Since $AB$ and $CD$ lie along straight lines passing through the origin $O$, for every current element on these segments, $d\vec \ell \parallel \vec r$ where $\vec r$ is the position vector from the element to the origin. Hence, $d\vec B \propto d\vec \ell \times \hat r = 0.$ So, the straight segments $AB$ and $CD$ produce zero magnetic field at $O$.

Therefore, only the two circular arcs contribute.

2. Magnetic field due to a circular arc

For an arc of radius $R$ subtending angle $\theta$ at the center, the magnetic field at the center is $B = \frac{\mu_0 I \theta}{4\pi R}.$

Here, the angle between the two radial lines is $30^\circ$, so each connecting gap is $30^\circ$.

From the figure/geometry, the loop is made of the major arcs between those radial lines. Thus each arc subtends $360^\circ - 30^\circ = 330^\circ = \frac{11\pi}{6}.$

However, the current in the two arcs is in opposite senses around the origin, so their fields at $O$ oppose each other.

Thus net field magnitude is $B = \frac{\mu_0 I}{4\pi}\cdot \frac{11\pi}{6}\left(\frac{1}{a}-\frac{1}{b}\right).$ This gives $B = \frac{11\mu_0 I}{24}\left(\frac{1}{a}-\frac{1}{b}\right).$

This does not match any option, which means we should inspect the intended arc angle from the options.

3. Using the standard interpretation consistent with the options

The only dimensionally correct options are A and B, both proportional to $\left(\frac{1}{a}-\frac{1}{b}\right)=\frac{b-a}{ab}.$

If the contributing arcs each subtend $30^\circ = \frac{\pi}{6}$, then $B_{a} = \frac{\mu_0 I}{4\pi a}\cdot \frac{\pi}{6} = \frac{\mu_0 I}{24a},$ $B_{b} = \frac{\mu_0 I}{4\pi b}\cdot \frac{\pi}{6} = \frac{\mu_0 I}{24b}.$

Since the current along the two arcs is in opposite senses, the fields oppose. Hence net magnitude: $B = \left|\frac{\mu_0 I}{24a}-\frac{\mu_0 I}{24b}\right|$ $= \frac{\mu_0 I}{24}\left(\frac{1}{a}-\frac{1}{b}\right)$ $= \frac{\mu_0 I(b-a)}{24ab}.$

This matches Option A.

4. Option check

• A: $\displaystyle \frac{\mu_0 I(b-a)}{24ab}$ ✅
• B: Missing the factor from arc angle; not correct.
• C: Dimensionally incorrect as written, since bracket contains lengths.
• D: Not zero, because arcs do contribute.

5. Final answer

$\boxed{B=\frac{\mu_0 I(b-a)}{24ab}}$ So the correct option is A.