Question

A proton with a speed of 2.20×106 m/s is shot into a region between two plates that are separated by a distance of 0.185 m. As the drawing shows, a magnetic field exists between the plates, and it is perpendicular to the velocity of the proton. What must be the magnitude of the magnetic field (in T), so the proton just misses colliding with the opposite plate?

what are the steps and formulas to acquire this answer?

Answer 1

We know that the proton cannot hit the other plate, which means the radius of its circular path cannot be larger than .185m.

If we use the equation:

r = mv/qB

Where r is the radius of its circular path, m is the mass of the particle, q is the charge of the particle, and B is the magnetic field, we rearrange for B, the magnetic field:

B = mv/QR

We now plug in our known values:

B = [(2.20 x 10⁶ m/s)(1.67 x 10^-27kg)] / [(1.602 x 10^-19C)(.185m)]

Where m_p = 1.67 x 10^-27kg and the charge of a proton is q = 1.602 x 10^-19C

We now solve:

B = .123967 T

If the magnetic field is any larger, the proton will hit the plate! I hope this makes sense!