To match the current configurations with the magnetic field at point $O$, we use standard results from Biot–Savart law.

1. Useful standard results

(i) Magnetic field at the center due to a circular arc

If a current $I$ flows in an arc of radius $r$ subtending angle $\theta$ at the center, then

$$B_{\text{arc}}=\frac{\mu_0 I\theta}{4\pi r}.$$

So:

• for a semicircle $(\theta=\pi)$,

$$B=\frac{\mu_0 I}{4r} = \frac{\mu_0 I}{4\pi r}\,\pi$$

• for a quarter circle $(\theta=\pi/2)$,

$$B=\frac{\mu_0 I}{8r}$$

(ii) Magnetic field due to a finite straight wire

At perpendicular distance $r$ from the wire,

$$B=\frac{\mu_0 I}{4\pi r}(\sin\theta_1+\sin\theta_2).$$

Special cases:

• semi-infinite wire:

$$B=\frac{\mu_0 I}{4\pi r}$$

• infinite wire:

$$B=\frac{\mu_0 I}{2\pi r}$$

In these figures, the total field at $O$ is obtained by adding contributions from arcs and straight segments.

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2. Match each configuration

From the given answer forms:

• I: $\frac{\mu_0 I}{4\pi r}(\pi+2)$
• II: $\frac{\mu_0 I}{4}\frac{I}{r}=\frac{\mu_0 I}{4r}$
• III: $\frac{\mu_0 I}{2\pi r}(\pi-1)=\frac{\mu_0 I}{4\pi r}(2\pi-2)$
• IV: $\frac{\mu_0 I}{4\pi r}(\pi+1)$

Now interpret these combinations:

Case corresponding to II

This is simply

$$\frac{\mu_0 I}{4r},$$

which is the magnetic field due to only a semicircular arc, with no net contribution from straight parts at $O$. Hence the configuration with only effective semicircle contribution must be matched to II. From the options, this corresponds to D $\to$ II in all viable answers.

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Case corresponding to IV

$$\frac{\mu_0 I}{4\pi r}(\pi+1)$$

This means:

• semicircle contribution: $\displaystyle \frac{\mu_0 I}{4\pi r}\pi$
• plus one semi-infinite straight wire contribution: $\displaystyle \frac{\mu_0 I}{4\pi r}$

So this configuration is semicircle + one effective straight contribution. Thus this matches C $\to$ IV.

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Case corresponding to I

$$\frac{\mu_0 I}{4\pi r}(\pi+2)$$

This means:

• semicircle contribution: $\displaystyle \frac{\mu_0 I}{4\pi r}\pi$
• plus two semi-infinite straight wire contributions: $2\times \displaystyle \frac{\mu_0 I}{4\pi r}$

So this is the configuration where the field at $O$ is due to a semicircle and two tangential semi-infinite wires. Thus this matches B $\to$ I.

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Case corresponding to III

$$\frac{\mu_0 I}{2\pi r}(\pi-1)=\frac{\mu_0 I}{4\pi r}(2\pi-2)$$

This corresponds to a combination where arc contribution dominates but one straight-wire contribution subtracts, giving net factor $(\pi-1)$ in the natural $\frac{\mu_0 I}{2\pi r}$ form. Thus the remaining configuration matches A $\to$ III.

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3. Final matching

Therefore,

$$A\to III,\quad B\to I,\quad C\to IV,\quad D\to II$$

This corresponds to Option C.

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4. Comparison with stored answer

Stored correct answer = C.

Our derived answer also = C. So the answer agrees.