The magnetic force (\( \vec{F}_m \)) experienced by a charged particle with charge (\( q \)) moving with velocity (\( \vec{v} \)) in a magnetic field (\( \vec{B} \)) is given by:
\[ \vec{F}_m = q (\vec{v} \times \vec{B}). \]
Inside a solenoid, the magnetic field (\( \vec{B} \)) is uniform and directed along the axis of the solenoid.
Thus, the magnetic force on the electron is:
\[ \vec{F}_m = 0. \]
Since the magnetic force on the electron is zero, there is no net force acting perpendicular to its motion. As a result:
The electron will not experience a magnetic force (B), and it will continue to move along the axis of the solenoid (C).


In the first configuration (1) as shown in the figure, four identical charges \( q_0 \) are kept at the corners A, B, C and D of square of side length \( a \). In the second configuration (2), the same charges are shifted to mid points C, E, H, and F of the square. If \( K = \frac{1}{4\pi \epsilon_0} \), the difference between the potential energies of configuration (2) and (1) is given by: