Question

A proton moves with a speed of 240 m/s in the +x-direction into a region of a 4.5-T uniform magnetic field directed $62^\circ$ above the +x-direction in the x,y-plane. Calculate the magnitude of the magnetic force on the proton.

Final Answer

$1.5\times 10^{-16}\textrm{ N}$

### Solution video

# OpenStax College Physics for AP® Courses, Chapter 22, Problem 8 (Test Prep for AP® Courses)

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Video Transcript

This is College Physics Answers with Shaun Dychko. A proton is moving in the positive x-direction with a speed of 240 meters per second and it's in the presence of a magnetic field which is directed 62 degrees above the positive x-axis and so this angle here is

*Θ*equals 62 degrees and it has a magnitude of 4.5 tesla and the question is what is the magnitude of the magnetic force on the proton? So the force is the charge multiplied by its speed times the magnetic field strength times*sin*of this angle between the field and the direction of the velocity. So because it's a proton, it has an elementary charge of 1.6 times 10 to the minus 19 coulombs times that by 240 meters per second times 4.5 tesla times*sin*62 degrees gives a force of 1.5 times 10 to the minus 16 newtons.