# What is the magnitude of the magnetic force on the proton?

+1 vote

A proton travels with a speed of 4.90  106 m/s at an angle of 63° with the direction of a magnetic field of magnitude 0.160 T in the positive x-direction.

(a) What is the magnitude of the magnetic force on the proton?
N

(b) What is the proton's acceleration?
m/s2
asked Feb 10, 2012 in Physics

+1 vote

The formula for Magnetic Force is:

Magnetic Force = q*v X B = q*v*B*sin(theta)   (*where q=charge, v=velocity, B=magnetic field)

the charge of a proton = 1.602*10^-19 C

v = 4.90  106 m/s

B = 0.160 T

theta = 63 degrees

Plugging in the formula we get:

Magnetic Force = (1.602*10^-19 C) * (4.90  106 m/s) X (0.160 T ) * sin(63) = 1.119*10^-13 N

Magnetic Force =  1.119*10^-13 N

Part B,

Force = mass* acceleration,

So divide answer from part A by the mass of a proton which is, 1.673*10^-27 kg

and so 1.119*10^-13 N/ 1.673*10^-27 kg = 6.689*10^13 m/s^2

answered Feb 18, 2012 by ~Expert~ (3,856 points)
selected Feb 18, 2012 by Joey33