An electron and a proton are moving under the influence of mutual forces. In calculating the change in the kinetic energy of the system during motion, one ignores the magnetic force of one on another. This is because

The correct answer is: the magnetic forces do no work on each particle.

To understand why we can ignore the magnetic forces when calculating the change in the kinetic energy of the system, let us analyze the nature of the magnetic force acting on a moving charged particle.

The magnetic force $\mathbf{F}$ acting on a particle with charge $q$ moving with velocity $\mathbf{v}$ in a magnetic field $\mathbf{B}$ is given by the Lorentz force formula:
$\mathbf{F}=q(\mathbf{v}\times \mathbf{B})$

By the definition of the cross product, the magnetic force vector $\mathbf{F}$ is always perpendicular to the velocity vector $\mathbf{v}$ of the particle:
$\mathbf{F}\cdot \mathbf{v}=0$

The rate at which work is done on the particle (power $P$) by this force is:
$P=\mathbf{F}\cdot \mathbf{v}=0$

Since the power is zero, the work done $W$ by the magnetic force on each particle over any time interval is also zero:
$W=\int \mathbf{F}\cdot d\mathbf{r}=\int (\mathbf{F}\cdot \mathbf{v})dt=0$

According to the work-energy theorem, the change in the kinetic energy of a particle is equal to the net work done on it. Since the magnetic force does no work on either the electron or the proton individually, it contributes nothing to the change in the kinetic energy of either particle or the system as a whole. Therefore, we can ignore the magnetic forces when calculating the change in the kinetic energy of the system.