- #1
rinarez7
- 27
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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.
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!
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!