magnetic force in physics

Question

A proton moves at a speed of 5.9 × 107 

m/s at right angles to a magnetic field with a 

magnitude of 0.095 T. 

Find the magnitude of the acceleration of 

the proton. 

Answer in units of m/s 

2 

.

Answer 1

Any charge that is moving in an external magnetic field experiences a force that is perpendicular to both the velocity vector and the magnetic field vector. This vector product is written as F_B = qv(cross)B (where F_B is the magnetic force vector experienced, q is the magnitude of the charge, v is the velocity vector, and B is the magnetic field vector). Since it's a vector product, we write the magnitude as |F_B| = qvB sinθ (θ = the angle between v and B vectors). Since this problem tells us that velocity is perpendicular to the magnetic field, sinθ is just 1, so it becomes qvB. Substituting the given information, 1.6x10^-19 C ^. 5.9x10⁷ m/s ^. 0.095 T = 8.9x10^-13 N for the force. Divide by mass 1.67x10^-27 kg to get acceleration = 5.4x10¹⁴ m/s².