Correct Option: (B) Zero

Step 1 (Governing Principle): [0.5 Mark]

The electric field $\vec{E}$ at the centre of a uniformly charged ring is determined by the principle of superposition. For every infinitesimal charge element $dq$ on the ring, there exists a diametrically opposite charge element $dq'$ of equal magnitude. The electric field produced by these two elements at the centre is given by:

$$d\vec{E} = \frac{1}{4\pi\varepsilon_0} \frac{dq}{R^2} \hat{r}$$

Step 2 (Mathematical Working): [0.25 Mark]

Consider two diametrically opposite elements of length $dl$ on the ring. The charge on each element is $dq = \lambda dl$. The electric field produced by these elements at the centre $O$ are equal in magnitude but opposite in direction:

$$\vec{E}_1 = \frac{1}{4\pi\varepsilon_0} \frac{\lambda dl}{R^2} \hat{r}$$ $$\vec{E}_2 = \frac{1}{4\pi\varepsilon_0} \frac{\lambda dl}{R^2} (-\hat{r})$$

The resultant field at the centre is the vector sum of all such pairs:

$$\vec{E}_{net} = \oint d\vec{E} = \int_{0}^{2\pi R} \frac{1}{4\pi\varepsilon_0} \frac{\lambda dl}{R^2} \hat{r}$$

Due to the symmetry of the circular ring, for every vector $\vec{E}$ at the centre, there is an equal and opposite vector $-\vec{E}$ resulting from the charge element on the opposite side.

Step 3 (Final Concluded Answer): [0.25 Mark]

Since the vector sum of all such diametrically opposite pairs cancels out completely:

$$\vec{E}_{net} = 0$$

Thus, the magnitude of the electric field at the centre is Zero.

💡 EL10 Concept Insight: Imagine the ring is like a tug-of-war game where every point on the ring is pulling the center toward itself; because the ring is perfectly circular and the charge is spread out evenly, every "pull" in one direction is perfectly cancelled out by an identical "pull" from the opposite side, leaving the center perfectly still.