1. Magnetic field inside a long solenoid

   Inside a long current-carrying solenoid, the magnetic field is uniform and directed along the axis of the solenoid.

   So, if the electron is projected along the axis, its velocity $\vec{v}$ is parallel to the magnetic field $\vec{B}$.

2. Magnetic force on a moving charge

   The magnetic force is given by

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

   Its magnitude is

   $$F = qvB\sin\theta$$

   where $\theta$ is the angle between $\vec{v}$ and $\vec{B}$.

3. Here $\theta = 0^\circ$

   Since the electron moves along the axis and the magnetic field is also along the axis,

   $$\theta = 0^\circ$$

   Hence,

   $$F = qvB\sin 0^\circ = 0$$

   Therefore, the electron experiences no magnetic force.

4. Conclusion about motion

   Since there is no force on the electron, there is no acceleration due to the magnetic field. So it will keep moving in a straight line with the same speed along the axis.

5. Option check

   • A: Incorrect. A helical path occurs when velocity has a component perpendicular to $\vec{B}$ as well as along $\vec{B}$. Here velocity is purely along $\vec{B}$.
   • B: Incorrect. Magnetic force cannot accelerate a particle along the field direction here; in fact force is zero.
   • C: Incorrect. Circular motion occurs when velocity is perpendicular to $\vec{B}$.
   • D: Correct. The electron continues with uniform velocity along the axis.

Final Answer