A proton travels along a circle of radius 1 m in a perpendicular magnetic field of strength 0.50 T . What would its speed be?

Step-by-step solution

The radius of the circular path of a charged particle in a perpendicular magnetic field is given by r = mv qB Rearranging for speed v , we get v = qBr m For a proton, charge q = 1.6 10⁻¹⁹ C and mass m = 1.67 10⁻²⁷ kg Given B = 0.50 T and r = 1 m Substituting the values: v = 1.6 10⁻¹⁹ 0.50 1 1.67 10⁻²⁷ v = 0.8 10⁻¹⁹ 1.67 10⁻²⁷ v 4.8 10^7 m s ⁻¹