A proton moves through a region containing a uniform electric field given by $=57\mathrm{.}0$ V$/$m and a uniform magnetic field $=(0\mathrm{.}200+0\mathrm{.}300+0\mathrm{.}400)$ T$.$ Determine the acceleration of the proton when it has a velocity $=231$ m$/$s$.$

Step $1$

The electric field in this problem is weak, the magnetic field is strong, and the velocity of the proton is small for a proton, on the order of a few hundred meters per second. Both fields contribute to the net force on the proton. This force is on the order of a fraction of a newton and will cause an acceleration of large magnitude. The acceleration due to gravity is negligible.

Step $2$

We will use the Lorentz force equation, which computes the vector addition of the forces on the particle. We will use Newton's second law with the particle under net force model.

Step $3$

Applying Newton's second law and the Lorentz force law, we have

$=$ m $=$ q $+$ q $.$

Solving for the acceleration gives

$=$

em

$+$

$,$

where e is the charge and m is the mass of the proton. For

$,$

we have the following determinant in which we have suppressed the SI units for this step.

$$

$23100$

$=+$