A charged particle moving in a magnetic field experiences a force equal to:
[tex]\vec{F}=q\vec{v}\times \vec{B}[/tex]
Thus, the magnitude of the force that the proton experiences is given by:
[tex]F=qvBsin\theta[/tex]
The magnetic field is perpendicular to the proton's velocity, therefore, we have [tex]\theta=90^\circ[/tex]. Replacing the given values, we obtain:
[tex]F=1.6*10^{-19}C(300\frac{m}{s})(1T)sin(90^\circ)\\F=4.8*10^{-17}N[/tex]
