The Lorentz force equation is given as:

$$\vec{F} = -q(\vec{v} \times \vec{B})$$

The electron is moving along the positive x-axis, so its velocity vector is $$\vec{v} = v_x \hat{i}$$. The magnetic field is applied parallel to the negative z-axis, so its magnetic field vector is $$\vec{B} = -B_z \hat{k}$$.

Now, we can calculate the cross product of the velocity and magnetic field vectors:

$$\vec{v} \times \vec{B} = (v_x \hat{i}) \times (-B_z \hat{k})$$

Using the cross product properties, we get:

$$\vec{v} \times \vec{B} = -v_x B_z (\hat{i} \times \hat{k})$$

The cross product of $$\hat{i}$$ and $$\hat{k}$$ is $$-\hat{j}$$, so:

$$\vec{v} \times \vec{B} = -v_x B_z (-\hat{j}) = v_x B_z \hat{j}$$

Since the electron has a negative charge, the magnetic force will be in the opposite direction:

$$\vec{F} = -(-e)(v_x B_z \hat{j}) = e(v_x B_z \hat{j})$$

As a result, the electron will experience a magnetic force along the negative y-axis.

Additionally, as mentioned earlier, when a charged particle moves through a magnetic field perpendicular to its velocity, it follows a circular path. In this case, the velocity of the electron is along the positive x-axis, and the magnetic field is along the negative z-axis, which are indeed perpendicular to each other. As a result, the electron will move along a circular path in the magnetic field.

Hence, the correct answer is:

(C) B and E only