A 4.9-MeV (kinetic energy) proton enters a 0.28-T field, in a plane perpendicular to the field. Part APart complete What is the radius of its path?

Explanation:

is the angle between the velocity and the magnetic field. So, the magnetic force on the proton is:

A charged particle describes a semicircle in a uniform magnetic field. Therefore, applying Newton's second law to uniform circular motion:

is the centripetal force and is defined as:

Here is the proton's speed and is the radius of the circular motion. Replacing this in (1) and solving for r:

Recall that 1 J is equal to , so:

We can calculate from the kinetic energy of the proton:

Finally, we calculate the radius of the proton path:

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