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}* \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(m)/(s))(1T)sin(90^\circ)\\F=4.8*10^(-17)N`