1. Angular momentum of a particle about the center of the circular path

For a particle moving in a circle, angular momentum is

$\vec L = \vec r \times \vec p = \vec r \times m\vec v$

where:

• $\vec r$ = position vector from the center,
• $\vec p = m\vec v$ = linear momentum.

2. Magnitude of angular momentum

In uniform circular motion, the particle moves with constant speed, so $|\vec v|$ is constant. Also, $|\vec r| = r$ is constant.

Since $\vec r$ is perpendicular to $\vec v$ in circular motion,

$L = mvr\sin 90^\circ = mvr$

So the magnitude remains constant.

3. Direction of angular momentum

The direction of $\vec L = \vec r \times m\vec v$ is perpendicular to the plane of motion, given by the right-hand rule.

As the particle keeps moving in the same circular plane, this perpendicular direction does not change.

So the direction also remains constant.

4. Checking options

• A: changes in magnitude but remains same in direction — False
• B: remains same in magnitude and direction — True
• C: remains same in magnitude but changes in the direction — False
• D: is zero — False

5. Final answer

Therefore, the angular momentum of a single particle moving with constant speed along a circular path remains same in magnitude and direction.