Proton in Magnetic field - am I correct?

AI Thread Summary
The discussion revolves around calculating the kinetic energy of a proton moving in a magnetic field. One participant presents a solution yielding approximately 1.3 x 10^5 eV, while another calculates around 1.17 x 10^5 eV. Key points of contention include the correct value for the charge of the proton and discrepancies in the energy conversion from joules to electron volts. The second participant points out a potential error in the first participant's calculations regarding the charge value used and the final energy conversion. The conversation emphasizes the importance of accuracy in physics calculations and peer review for verification.
Poobel
Messages
25
Reaction score
0
Proton in Magnetic field -- am I correct?

One of my Physics ISU questions is as follows:
- A proton moves in a circular path perpendicular to a 1.10 T magnetic field. The radius of the path is 4.5 cm. calculate energy of the proton in eV

Well i had hard time doing it but here we go: my solution

mV^2/r=QrB V=QrB/m

V= (1.6x10^-16) x o.o45 m x 1.10 T) / 1.67x10^-27 kg

V= 4.7x10^6 m/s

Ek= 1/2mV^2 = 1/2 x 1.67x10^-27 kg x (4.7x10^6 m/s)^2
= 1.8x10^-14 J

Ek= 1.87x10^-14J x eV/1.6x10^-19J = (1.87x10^-14J)/ (1.6x10^-19J)

eV=1.3x10^5 eV=1.3x10^5



Fellow physisists is my solution correct?

Thank you
 
Physics news on Phys.org
Hmm...I got close to 1.17*10^5 eV

Here's how I did it:

We want to know the kinetic energy of the proton, so:

E_{k}=\frac{1}{2}mv^2

We know the mass of a proton is 1.67*10^-27, so the variable we need is the velocity at which it is travelling.

As this proton is moving in a circle in a magnetic field,

F_m=F_c (where F_m is the magnetic force and F_c is the centripetal force

representing these with equations, we have

qvB=\frac{mv^2}{r}

now, we isolate v, since that is what we are trying to get

v=\frac{qBr}{m}

substituting into the kinetic energy equation, we get

E_{k}=\frac{1}{2}m(\frac{qBr}{m})^2

which reduces to

E_{k}=\frac{(qBr)^2}{2m}

substituting in,

E_{k}=\frac{((1.6*10^-19)(1.10)(0.045))^2}{2(1.67*10^-27)}

which gets 1.878*10^-14 J

since 1 eV = 1.6*10^-19 J, the answer in eV is

(1.878*10^-14)/(1.6*10^-19) = 117,375 eV (1.17*10^5)

There are two things about yours that confuse me. First off, the magnitude of the charge on a proton is the elementary charge, which is 1.6*10^-19, whereas you have it as 1.6*10^-16...and yet you get the same energy in joules. Then at the end when you do your conversion from joules to electron volts, there seems to be a discrepency of some kind, because (1.87*10^-14)/(1.6*10^-19) does not equal 1.3*10^5. Also, I would have rounded to 1.88*10^5 if I were rounding. Apart from that the solution the same as mine.
 
Last edited:
Ok I'll look more into it

thanks a bunch, help is always appreciated

BTW that -16... I just copied it wrong from my notes:smile:

Hmm also you must've read my first post wrong since

Ek= 1.87x10^-14J x eV/1.6x10^-19J = (1.87x10^-14J)/ (1.6x10^-19J)

eV=1.3x10^5 eV=1.3x10^5
 
Last edited:
I multiplied the values first without the error limit. Got 19.38. rounded it off to 2 significant figures since the given data has 2 significant figures. So = 19. For error I used the above formula. It comes out about 1.48. Now my question is. Should I write the answer as 19±1.5 (rounding 1.48 to 2 significant figures) OR should I write it as 19±1. So in short, should the error have same number of significant figures as the mean value or should it have the same number of decimal places as...
Thread 'A cylinder connected to a hanging mass'
Let's declare that for the cylinder, mass = M = 10 kg Radius = R = 4 m For the wall and the floor, Friction coeff = ##\mu## = 0.5 For the hanging mass, mass = m = 11 kg First, we divide the force according to their respective plane (x and y thing, correct me if I'm wrong) and according to which, cylinder or the hanging mass, they're working on. Force on the hanging mass $$mg - T = ma$$ Force(Cylinder) on y $$N_f + f_w - Mg = 0$$ Force(Cylinder) on x $$T + f_f - N_w = Ma$$ There's also...

Similar threads

Back
Top