1. A singly charged positive ion has a mass = 1.08e-26 kg After being accelerated through a potential difference of 240 V the ion enters a magnetic field of 0.595 T in a direction perpendicular to the field. The charge on the ion is 1.602 e-19 C. Find the radius of the ion's path in the field. Answer in units of cm. 2. m= 1.08e-26 kg potential difference= 240 V B= 0.595 T q= 1.602e-19 sin theta= 1 (enters in direction perpendicular to the field) radius= mv/ qB 3. I am confused on exactly how to calculate velocity and the relevance of the potential difference. But, I tried: F=qvBsin theta= ma F=qvB= ma 1.602e-19(v) (0.595T)= 1.08e-26(-9.81m/s^2) v= 1.112e-6 m/s I then plugged this into r= mv/ qB r= 1.08e-26kg (1.112e-6 m/s )/ 1.602e-19 (0.595T) and calculated r= 1.8997 e-11 cm, but this incorrect. What am I missing?