Question

A proton enters a uniform magnetic field of strength 1 T at 300 m/s. The magnetic field is oriented perpendicular to the proton’s velocity. What is the magnitude of the force that the proton experiences while it moves through the magnetic field?

Answer 1

A charged particle moving in a magnetic field experiences a force equal to:

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

Thus, the magnitude of the force that the proton experiences is given by:

$F=qvBsin\theta$

The magnetic field is perpendicular to the proton's velocity, therefore, we have $\theta=90^\circ$. Replacing the given values, we obtain:

$F=1.6*10^{-19}C(300\frac{m}{s})(1T)sin(90^\circ)\\F=4.8*10^{-17}N$