Finding mass of charged particle in circular motion through magnetic field

[rinarez7]

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.

Homework 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!

 

[LowlyPion]

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.

Homework 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!

What is the E field? I don't see that.
http://hyperphysics.phy-astr.gsu.edu/hbase/magnetic/magfor.html#c2

Don't you have a charge that is moving in a circle in a B field.

|qV X B| = mV²/R

And you know that 1/2mV² = .115 J