An electron moving in a circular orbit of radius r makes n rotations per second. The magnetic field produced at the centre has magnitude:

A. (μ₀ n² e) / r
B. (μ₀ n e) / 2r
C. (μ₀ n e) / (2π r)
D. Zero

To determine the magnetic field produced at the center of a circular orbit by an electron, we can use the concept of the magnetic field produced by a current-carrying loop.

**Given:**
- Radius of the circular orbit: \( r \)
- The electron makes \( n \) rotations per second (which is the frequency of rotation).
- The charge of an electron: \( e \).

**Step-by-step solution:**

1. **Calculate the current** produced by the moving electron:

When an electron moves in a circular orbit, it creates a current. The current \( I \) is defined as the charge passing through a point per unit time. Since the electron is moving in a circle and making \( n \) rotations per second, the current due to this motion is:

\[
I = \text{(charge per rotation)} \times \text{(number of rotations per second)}
\]
\[
I = e \times n
\]

2. **Magnetic field due to a current-carrying loop:**

The magnetic field \( B \) at the center of a circular loop carrying current \( I \) is given by Ampere's law:

\[
B = \frac{\mu_0 I}{2r}
\]

where \( \mu_0 \) is the permeability of free space and \( r \) is the radius of the loop.

3. **Substitute the current \( I = en \) into the formula for \( B \):**

\[
B = \frac{\mu_0 (en)}{2r}
\]

This matches option B. Thus, the correct answer is:

\[
\boxed{B. \frac{\mu_0 ne}{2r}}
\]