A proton moves perpendicular to a uniform magnetic field B v

A proton moves perpendicular to a uniform magnetic field B vector at a speed of 1.50 * 18^2 m/s and experiences and declaration of 1.90 * 10^13 m/s^2 in the positive x - direction of the field.

Solution

B = F / q v = m a / q v

= (1.67 * 10^-27 * 1.90 * 10¹³) / (1.6 * 10^-19 * 1.50 * 10⁷)

B = 0.013 T

The right hand rule shows that B must be in the -y direction to yield a force in the +x direction when v is in the +z direction.