Mass of charged particle in 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

 

[nlightner]

I would like to thank you for sharing your work and thought process. It seems that you have made a small error in your calculations. Instead of using the kinetic energy formula, KE= mv^2/2, you should use the formula for centripetal force, Fc= mv^2/r, since the particle is moving in a circular path. This will give you the correct value for the mass.

Also, it is important to note that the mass of the charged particle may be very small and may require the use of scientific notation to accurately represent it. In this case, the mass is 3 x 10^-9 kg, which is a very small value.

Additionally, it is always a good practice to double-check your calculations and make sure all units are consistent throughout your work. Keep up the good work and always strive for accuracy in your scientific calculations!