Angular momentum of a single particle moving with constant speed along circular path:

We start with the definition of angular momentum of a single particle about the centre of the circular path.

By definition, $$\vec L = \vec r \times \vec p,$$ where $$\vec r$$ is the position vector of the particle with respect to the centre, and $$\vec p = m\vec v$$ is its linear momentum.

For uniform circular motion the radius is constant, so $$|\vec r| = R,$$ and the speed is constant, so $$|\vec v| = v.$$ The position vector $$\vec r$$ is always radial, while the velocity $$\vec v$$ is always tangential, making the angle between them $$90^\circ.$$

Using the formula for the magnitude of a cross-product, $$|\vec A \times \vec B| = |\vec A|\,|\vec B|\,\sin\theta,$$ we substitute $$\vec A = \vec r,\ \vec B = \vec p,\ \theta = 90^\circ.$$ This gives

$$|\vec L| = |\vec r|\,|\vec p|\,\sin 90^\circ = R\,(mv)\,(1) = m v R.$$

The quantities $$m,\,v,$$ and $$R$$ are each constant during the motion, so the magnitude $$|\vec L|$$ remains fixed.

The direction of $$\vec L$$ is given by the right-hand rule applied to $$\vec r \times \vec p.$$ Because $$\vec r$$ is always in the plane of the circle and $$\vec p$$ is always in that same plane but perpendicular to $$\vec r,$$ the cross-product is always perpendicular to the plane of the circle. If the motion is anticlockwise, $$\vec L$$ points steadily “out of the page”; if the motion is clockwise, it points steadily “into the page.” In either case the direction is fixed and does not change as the particle moves around.

We can confirm this with torque. Torque about the centre is $$\vec\tau = \vec r \times \vec F.$$ The only force needed for uniform circular motion is the centripetal force $$\vec F_c = -\dfrac{mv^2}{R}\,\hat r,$$ directed along $$-\hat r.$$ Since $$\vec r$$ is parallel to $$\hat r,$$ their cross-product is zero:

$$\vec\tau = \vec r \times \vec F_c = 0.$$

With zero external torque, the rotational analogue of Newton’s second law, $$\vec\tau = \dfrac{d\vec L}{dt},$$ immediately gives $$\dfrac{d\vec L}{dt} = 0.$$ Therefore the vector $$\vec L$$ is completely constant—both its magnitude and its direction stay unchanged.

Hence, the correct answer is Option B.