Question

A proton moves at 8.00 ✕ 10⁷ m/s perpendicular to a magnetic field. The field causes the proton to travel in a circular path of radius 0.750 m. What is the field strength (in T)?

Answer 1

When a charged particle enters perpendicular to a magnetic field, it executes circular motion. The centripetal force required for the circular motion is provided by the magnetic foce. Therefore,

where v is the velocity of the charged particle, m is the mass of the charged particle, q is the mass of the charged particle, B is the magnetic field and r is the radius of the charged particle.

Given v = 8 x 10^7 m/s and r = 0.750m. For a proton m = 1.67 x 10^-27kg and q = 1.6 x 10^-19C. Therefore,

So the magnetic field strength is 1.11T.