A proton is moving at 2.1 x 10 m/s clockwise through a constant and perpendicular magnetic field. The radius of curvature through the field is 0.6 meters. What is the strength of the magnetic field?

For this question the magnetic force provides the force required for the circular motion

Equating

Magnetic force = centripetal force

Bqvsinx = mv^2/r

as magnetic field is perpendicular x=90

Bqvsin90 = mv^2/r

Bq = mv/r

B = mv/rq

Then replace r = 0.6, v = 2.1ms and mp from formula sheet and you can obtain B which is magnetic field intensity