Answer to Question #267277 in Electric Circuits for zuzu

Question

a) A proton in a particle accelerator has a speed of 4.0 x 10⁶ m/s. The proton encounters a magnetic field whose magnitude is 0.50T and whose direction makes an angle of 20o with respect to the proton's velocity. Find the magnitude and direction of the magnetic force on the proton. 

Answer 1

force on the proton:

$\vec{F_B}=q(\vec{v}\times\vec{B})$

$F=|q|vBsin\theta=1.6\cdot10^{-19}\cdot4\cdot10^6\cdot0.5sin20\degree=1.09\cdot10^{-13}$ N

direction is upward