Magnetic Force

Question

What is the magnetic force on a particle that has –3.2 × 10–19 C of charge and is moving at 2.5 × 107 m/s to the right through a magnetic field that is 0.24 T and pointing away from you? Specify both magnitude and direction in your answer.

Answer 1

Magnitude:

 

F = q v x B

 

Since v and B are perpendicular in this problem,

 

F = qvB = (3.2 x 10^-19)(2.5 x 10^7)(0.24) N = 1.92 x 10^-12 N

 

Direction:

 

v points to the right, B points into the page, so by the right hand rule vxB is in the plane of the page pointing toward the top of the page.  The charge is negative, which reverses the direction.  Hence qvxB is in the plane of the page pointing toward the bottom, and that is the direction of the magnetic force.