A proton travels with a speed of 284 x 10 ms at an angle of

A proton travels with a speed of 2.84 x 10 m/s at an angle of 32.1degree with a magnetic field of 0.365 T pointed in the y direction. The charge of proton Ls 1.60218 x 10^-19 C. What is the magnitude of the magnetic force on the proton? Answer in units of N. The mass of proton is 1.67262 x 10^-27 kg What is the proton\'s acceleration? Answer in units of m/s^2.

Solution

F = q v B sin theta

= 1.60 * 10^-19 * 2.84 * 10⁶ * 0.365 * sin 32.1

F = 8.81 * 10^-14 N

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a = F / m

= 8.81 * 10^-14/ 1.67 * 10^-27

a = 5.27 * 10¹³ m/s²