1. A 1.56024 μC charged particle with a kinetic energy of 0.115905 J is placed in a uniform magnetic field of magnitude 0.150267 T. If the particle moves in a circular path of radius 3.13685 m, find its mass. Answer in units of kg. 2. Relevant equations Fe= Fm Fe= qE KE=mv^2/2 Fm= qvB= mv^2/r 3. I decided to first find E: E =Ke (q/r^2)= 8.98e10 (1.5602e-6 C/3.13685^2)=1425.455 Then I solved for Fe= qE= 1425.455 (1.5602e-6)= 2.2239e-3= Fm Then I used Fm= qvB and solved for v = 9503.84615 m/s Then I used KE = mv^2/2 to sove for m = KE(2)/ v^2= (0.115905)(2)/ (9503.84615 m/s )^2=3e-9 kg=m But this isn't correct. What am I missing here? Thank you in advance for any help!